3D Geometry · Mathematics · JEE Main

Let the foot of perpendicular from the point $(\lambda, 2,3)$ on the line $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$ be the point ( $1, \mu, 2$ ). Then the distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $\frac{x-\lambda}{2}=\frac{y-\mu}{3}=\frac{z+5}{6}$ is equal to :

The shortest distance between the lines $\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$ and $\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$ is:

Let the image of the point $\mathrm{P}(1,6, a)$ in the line $\mathrm{L}: \frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}, b>0$, be $\left(\frac{a}{3}, 0, a+c\right)$. If $\mathrm{S}(\alpha, \beta, \gamma), \alpha>0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2 \sqrt{14}$, then $\alpha+\beta+\gamma$ is equal to:

Let a line L be perpendicular to both the lines $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$.

If $\theta$ is the acute angle between the lines L and $\mathrm{L}_3: \frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}$, then $\tan \theta$ is equal to:

Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ and are at a distance of 6 units from $R(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta P Q R$, then $\alpha+\beta+\gamma$ is equal to :

If the distance of the point $(a, 2,5)$ from the image of the point $(1,2,7)$ in the line $\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{2}$ is 4 , then the sum of all possible values of $a$ is equal to :

The square of the distance of the point $\mathrm{P}(5,6,7)$ from the line $\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}$ is equal to:

$\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0$ and $\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})$ from the origin is :

The shortest distance between the lines

$$ \vec{r}=\left(\frac{1}{3} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\frac{8}{3} \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) $$

and $\vec{r}=\left(-\frac{2}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{k}}\right)+\mu(\hat{\mathrm{j}}-\hat{\mathrm{k}}), \lambda, \mu \in \mathbb{R}$, is:

If $\left(2 \alpha+1, \alpha^2-3 \alpha, \frac{\alpha-1}{2}\right)$ is the image of $(\alpha, 2 \alpha, 1)$ in the line $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}$, then the possible value(s) of $\alpha$ is (are)

A line with direction ratios $1,-1,2$ intersects the lines $\frac{x}{2}=\frac{y}{3}=\frac{z+1}{3}$ and $\frac{x+1}{-1}=\frac{y-2}{1}=\frac{z}{4}$ at the points P and Q , respectively. If the length of the line segment PQ is $\alpha$, then $225 \alpha^2$ is equal to:

The square of the distance of the point $(-2,-8,6)$ from the line $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}$ along the line $\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}$ is equal to:

Let the point A be the foot of perpendicular drawn from the point P$(a, b, 0)$ on the line

$$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}.$$

If the midpoint of the line segment PA is $$\left(0, \frac{3}{4}, -\frac{1}{4}\right),$$ then the value of $a^2 + b^2 + \alpha^2$ is equal to :

If the point of intersection of the lines $ \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} $ and $ \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} $ lies on xy-plane, then the value of $a+b$ is:

Let a line L passing through the point (1, 1, 1) be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line L, then the value of $34(a + b + c)$ is :

Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line $\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$. Then the distance of Q from the line $\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$ is

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\sqrt{2}$, is

Let the direction cosines of two lines satisfy the equations : $4 l+m-n=0$ and $2 m n+10 n l+3 l m=0$.

Then the cosine of the acute angle between these lines is :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. The coordinates of A and $B$ are $(1,6,3)$ and $(4,9, \alpha)$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $\triangle A B C$ is :

Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L , whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7 . Then $\sum\limits_{(a, b, c) \in S}(a+b+c)$ is equal to :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, b, \mathrm{c})$, then $a^2+b^2+c^2$ is equal to

Let the line L pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :

Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $\left|\overrightarrow{CD}\right|^2$ is equal to:

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

If the equation of the line passing through the point $ \left( 0, -\frac{1}{2}, 0 \right) $ and perpendicular to the lines $ \vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right) $ and $ \vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right) $ is $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, then $ a+b+c+d $ is equal to :

Consider the lines L₁: x - 1 = y - 2 = z and L₂: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L₁ and L₂ be Q and R respectively. If the area of the triangle PQR is A, then 4A² is equal to :

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then, which of the following points lies on the line $L$ ?

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Let A be the point of intersection of the lines $\mathrm{L}_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $\mathrm{L}_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively such that $A B=A C=\sqrt{15}$. Then the square of the area of the triangle $A B C$ is :

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b},(\mathrm{a}<\mathrm{b})$. Then the length of the latus rectum of the ellipse $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ is :

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2 \sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ is equal to

Let a line passing through the point $(4,1,0)$ intersect the line $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $\mathrm{L}_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|$ is equal to

Line $L_1$ passes through the point $(1,2,3)$ and is parallel to $z$-axis. Line $L_2$ passes through the point $(\lambda, 5,6)$ and is parallel to $y$-axis. Let for $\lambda=\lambda_1, \lambda_2, \lambda_2<\lambda_1$, the shortest distance between the two lines be 3 . Then the square of the distance of the point $\left(\lambda_1, \lambda_2, 7\right)$ from the line $L_1$ is

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, \mathrm{QR}=5$ and the coordinates of the point $P$ be $(0,2,3)$. If the area of the triangle $P Q R$ is $\frac{m}{n}$ then :

Let $A B C D$ be a tetrahedron such that the edges $A B, A C$ and $A D$ are mutually perpendicular. Let the areas of the triangles $\mathrm{ABC}, \mathrm{ACD}$ and ADB be 5,6 and 7 square units respectively. Then the area (in square units) of the $\triangle B C D$ is equal to :

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} $ and $ \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} $. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$.
Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to :

Let $\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines.

Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $\mathrm{L}_1$, then $|5 \alpha-11 \beta-8 \gamma|$ equals :

The square of the distance of the point $ \left( \frac{15}{7}, \frac{32}{7}, 7 \right) $ from the line $ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} $ in the direction of the vector $ \hat{i} + 4\hat{j} + 7\hat{k} $ is:

Let $\mathrm{A}(x, y, z)$ be a point in $x y$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and $(0,0,1)$.

Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements

(S1) : $\triangle \mathrm{ABC}$ is an isosceles right angled triangle, and

(S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$,

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

Let in a $\triangle A B C$, the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle A B C$ is:

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is

If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$, where $m$, $n$ are coprime numbers, then $m+n$ is equal to :

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is :

Let P be the foot of the perpendicular from the point $\mathrm{Q}(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $(3,-2,1)$, is

Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3 \hat{i}+2 \hat{j}+2 \hat{k}$. If the distance of the point Q from the point $\mathrm{R}(1,3,3)$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :

The perpendicular distance, of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $\mathrm{P}(2,-10,1)$, is :

Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $\mathrm{L}_1$ and $\mathrm{L}_2$ ?

Consider the line $$\mathrm{L}$$ passing through the points $$(1,2,3)$$ and $$(2,3,5)$$. The distance of the point $$\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$$ from the line $$\mathrm{L}$$ along the line $$\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$$ is equal to

The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:

Let the line $$\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$$. Then which of the following points lies on $$\mathrm{L}$$ ?

If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :

Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$x y$$-plane is the point $$Q$$. Let $$O P=\gamma$$; the angle between $$O Q$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$O P$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is

If the shortest distance between the lines

$$\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R} \end{array}$$

is $$\frac{m}{\sqrt{n}}$$, where $$\operatorname{gcd}(m, n)=1$$, then the value of $$m+n$$ equals

Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equal to :

If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is

The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is

Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha+\beta+\gamma$$ is equal to :

If the line $$\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$$ makes a right angle with the line $$\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$$, then $$4 \lambda+9 \mu$$ is equal to :

Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :

Let $$\mathrm{P}$$ be the point of intersection of the lines $$\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$$ and $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$$. Then, the shortest distance of $$\mathrm{P}$$ from the line $$4 x=2 y=z$$ is

Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2+\beta^2+\gamma^2$$ is equal to :

Let $$(\alpha, \beta, \gamma)$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Then, $$2 \alpha+3 \beta+4 \gamma$$ is equal to

The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is

Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,

$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$

be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then, the point which lies on $$L_3$$ is

Let $$(\alpha, \beta, \gamma)$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$$. Then $$19(\alpha+\beta+\gamma)$$ is equal to :

Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$A B C D$$. If the diagonal $$\overrightarrow{\mathrm{BD}}=\hat{i}+2 \hat{j}+3 \hat{k}$$, then the area of the parallelogram is equal to :

Let $$\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)$$ and $$\mathrm{R}(7,3,2)$$ be the vertices of $$\triangle \mathrm{PQR}$$. Then, the angle $$\angle \mathrm{QPR}$$ is

Let $$O$$ be the origin and the position vectors of $$A$$ and $$B$$ be $$2 \hat{i}+2 \hat{j}+\hat{k}$$ and $$2 \hat{i}+4 \hat{j}+4 \hat{k}$$ respectively. If the internal bisector of $$\angle \mathrm{AOB}$$ meets the line $$\mathrm{AB}$$ at $$\mathrm{C}$$, then the length of $$O C$$ is

Let $$P Q R$$ be a triangle with $$R(-1,4,2)$$. Suppose $$M(2,1,2)$$ is the mid point of $$\mathrm{PQ}$$. The distance of the centroid of $$\triangle \mathrm{PQR}$$ from the point of intersection of the lines $$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$$ and $$\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$$ is

Let the image of the point $$(1,0,7)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and making angles $$\frac{2 \pi}{3}$$ and $$\frac{3 \pi}{4}$$ with $$y$$-axis and $$z$$-axis respectively and an acute angle with $$x$$-axis ?

The line, that is coplanar to the line $$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$$, is :

The plane, passing through the points $$(0,-1,2)$$ and $$(-1,2,1)$$ and parallel to the line passing through $$(5,1,-7)$$ and $$(1,-1,-1)$$, also passes through the point :

Let $$\mathrm{N}$$ be the foot of perpendicular from the point $$\mathrm{P}(1,-2,3)$$ on the line passing through the points $$(4,5,8)$$ and $$(1,-7,5)$$. Then the distance of $$N$$ from the plane $$2 x-2 y+z+5=0$$ is :

Let the equation of plane passing through the line of intersection of the planes $$x+2 y+a z=2$$ and $$x-y+z=3$$ be $$5 x-11 y+b z=6 a-1$$. For $$c \in \mathbb{Z}$$, if the distance of this plane from the point $$(a,-c, c)$$ is $$\frac{2}{\sqrt{a}}$$, then $$\frac{a+b}{c}$$ is equal to :

The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$$ and $$\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$$ is :

Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to

Let the plane P: $$4 x-y+z=10$$ be rotated by an angle $$\frac{\pi}{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\mathrm{P}$$, then $$35 \alpha$$ is equal to :

Let the line passing through the points $$\mathrm{P}(2,-1,2)$$ and $$\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\mathrm{R}$$. Then the distance of the point $$\mathrm{R}$$ from the plane $$x+2 y+3 z+2=0$$ measured parallel to the line $$\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$$ is equal to :

Let P be the plane passing through the points $$(5,3,0),(13,3,-2)$$ and $$(1,6,2)$$. For $$\alpha \in \mathbb{N}$$, if the distances of the points $$\mathrm{A}(3,4, \alpha)$$ and $$\mathrm{B}(2, \alpha, a)$$ from the plane P are 2 and 3 respectively, then the positive value of a is :

Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{P}(2,3,5)$$ in the plane $$2 x+y-3 z=6$$. Then $$\alpha+\beta+\gamma$$ is equal to :

If equation of the plane that contains the point $$(-2,3,5)$$ and is perpendicular to each of the planes $$2 x+4 y+5 z=8$$ and $$3 x-2 y+3 z=5$$ is $$\alpha x+\beta y+\gamma z+97=0$$ then $$\alpha+\beta+\gamma=$$

Let the image of the point $$\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$$ and $$\mathrm{C}(0,5,1)$$ be $$\mathrm{Q}(\alpha, \beta, \gamma)$$. Then $$\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$$ is equal to :

Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :

The shortest distance between the lines $${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$$ and $${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$$ is :

Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :

Let P be the point of intersection of the line $${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :

For $$\mathrm{a}, \mathrm{b} \in \mathbb{Z}$$ and $$|\mathrm{a}-\mathrm{b}| \leq 10$$, let the angle between the plane $$\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}$$ and the line $$l: x-1=\mathrm{a}-y=z+1$$ be $$\cos ^{-1}\left(\frac{1}{3}\right)$$. If the distance of the point $$(6,-6,4)$$ from the plane P is $$3 \sqrt{6}$$, then $$a^{4}+b^{2}$$ is equal to :

Let $$\mathrm{P}$$ be the plane passing through the line

$$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$$ and the point $$(2,4,-3)$$.

If the image of the point $$(-1,3,4)$$ in the plane P

is $$(\alpha, \beta, \gamma)$$ then $$\alpha+\beta+\gamma$$ is equal to :

The shortest distance between the lines $$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$ is :

If the equation of the plane containing the line

$$x+2 y+3 z-4=0=2 x+y-z+5$$ and perpendicular to the plane

$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$

is $a x+b y+c z=4$, then $$(a-b+c)$$ is equal to :

A plane P contains the line of intersection of the plane $$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$$ and $$\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$$. If $$\mathrm{P}$$ passes through the point $$(0,2,-2)$$, then the square of distance of the point $$(12,12,18)$$ from the plane $$\mathrm{P}$$ is :

Let the line $$\mathrm{L}$$ pass through the point $$(0,1,2)$$, intersect the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and be parallel to the plane $$2 x+y-3 z=4$$. Then the distance of the point $$\mathrm{P}(1,-9,2)$$ from the line $$\mathrm{L}$$ is :

If the equation of the plane passing through the line of intersection of the planes $$2 x-y+z=3,4 x-3 y+5 z+9=0$$ and parallel to the line $$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$$ is $$a x+b y+c z+6=0$$, then $$a+b+c$$ is equal to :

One vertex of a rectangular parallelopiped is at the origin $$\mathrm{O}$$ and the lengths of its edges along $$x, y$$ and $$z$$ axes are $$3,4$$ and $$5$$ units respectively. Let $$\mathrm{P}$$ be the vertex $$(3,4,5)$$. Then the shortest distance between the diagonal OP and an edge parallel to $$\mathrm{z}$$ axis, not passing through $$\mathrm{O}$$ or $$\mathrm{P}$$ is :

Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $$\mathrm{d^{2}}$$ is equal to :

The shortest distance between the lines

$${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$$ and

$${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$$ is :

Let the image of the point $$P(2,-1,3)$$ in the plane $$x+2 y-z=0$$ be $$Q$$.

Then the distance of the plane $$3 x+2 y+z+29=0$$ from the point $$Q$$ is :

Let the shortest distance between the lines

$$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$$ and

$$L_{1}: x+1=y-1=4-z$$ be $$2 \sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on $$L$$,

then which of the following is NOT possible?

The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :

The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to :

If the lines $${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$$ and $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$$ intersect at the point P, then the distance of the point P from the plane $$z = a$$ is :

The shortest distance between the lines $${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$$ and $${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$$ is :

The foot of perpendicular of the point (2, 0, 5) on the line $${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$$ is ($$\alpha,\beta,\gamma$$). Then, which of the following is NOT correct?

The shortest distance between the lines $$x+1=2y=-12z$$ and $$x=y+2=6z-6$$ is :

The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :

Consider the lines $$L_1$$ and $$L_2$$ given by

$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.

A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is

If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\alpha,\beta,\gamma$$), then $$\alpha+\beta+\gamma$$ is equal to :

Let the plane containing the line of intersection of the planes

P₁ : $$x+(\lambda+4)y+z=1$$ and

P₂ : $$2x+y+z=2$$

pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of

the point (2$$\lambda,\lambda,-\lambda$$) from the plane P₂ is :

The distance of the point (7, $$-$$3, $$-$$4) from the plane passing through the points (2, $$-$$3, 1), ($$-$$1, 1, $$-$$2) and (3, $$-$$4, 2) is :

The distance of the point ($$-1,9,-16$$) from the plane

$$2x+3y-z=5$$ measured parallel to the line

$${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$ is :

Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of $$\triangle P Q R$$ is equal to :

If $$(2,3,9),(5,2,1),(1, \lambda, 8)$$ and $$(\lambda, 2,3)$$ are coplanar, then the product of all possible values of $$\lambda$$ is:

If the foot of the perpendicular from the point $$\mathrm{A}(-1,4,3)$$ on the plane $$\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$$, is $$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $$3,-1,-4$$, is equal to :

Let the lines

$$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$$ and

$$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$$ be coplanar

and $$\mathrm{P}$$ be the plane containing these two lines.

Then which of the following points does NOT lie on P?

A plane P is parallel to two lines whose direction ratios are $$-2,1,-3$$ and $$-1,2,-2$$ and it contains the point $$(2,2,-2)$$. Let P intersect the co-ordinate axes at the points $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ making the intercepts $$\alpha, \beta, \gamma$$. If $$\mathrm{V}$$ is the volume of the tetrahedron $$\mathrm{OABC}$$, where $$\mathrm{O}$$ is the origin, and $$\mathrm{p}=\alpha+\beta+\gamma$$, then the ordered pair $$(\mathrm{V}, \mathrm{p})$$ is equal to :

The foot of the perpendicular from a point on the circle $$x^{2}+y^{2}=1, z=0$$ to the plane $$2 x+3 y+z=6$$ lies on which one of the following curves?

If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2 \sqrt{6}$$ units and $$Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$$ is equal to :

If the line of intersection of the planes $$a x+b y=3$$ and $$a x+b y+c z=0$$, a $$>0$$ makes an angle $$30^{\circ}$$ with the plane $$y-z+2=0$$, then the direction cosines of the line are :

If the plane $$P$$ passes through the intersection of two mutually perpendicular planes $$2 x+k y-5 z=1$$ and $$3 k x-k y+z=5, k<3$$ and intercepts a unit length on positive $$x$$-axis, then the intercept made by the plane $$P$$ on the $$y$$-axis is :

A vector $$\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\hat{i}, \hat{i}+\hat{j}$$ and the plane determined by the vectors $$\hat{i}-\hat{j}, \hat{i}+\hat{k}$$. The obtuse angle between $$\vec{a}$$ and the vector $$\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$$ is :

The length of the perpendicular from the point $$(1,-2,5)$$ on the line passing through $$(1,2,4)$$ and parallel to the line $$x+y-z=0=x-2 y+3 z-5$$ is :

A plane $$E$$ is perpendicular to the two planes $$2 x-2 y+z=0$$ and $$x-y+2 z=4$$, and passes through the point $$P(1,-1,1)$$. If the distance of the plane $$E$$ from the point $$Q(a, a, 2)$$ is $$3 \sqrt{2}$$, then $$(P Q)^{2}$$ is equal to :

The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :

Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $$(2,-5,11)$$, then $$\mathrm{d}^{2}$$ is equal to :

The distance of the point (3, 2, $$-$$1) from the plane $$3x - y + 4z + 1 = 0$$ along the line $${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$$ is equal to :

Let $${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\in$$ R. The shortest distance of the plane from the origin is :

Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $$\overrightarrow r = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right),\,\lambda \in R$$. Then, which of the following points lies on T?

If the mirror image of the point (2, 4, 7) in the plane 3x $$-$$ y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to :

Let the plane ax + by + cz = d pass through (2, 3, $$-$$5) and is perpendicular to the planes
2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :

If two distinct point Q, R lie on the line of intersection of the planes $$ - x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$ and $$PQ = PR = \sqrt {18} $$ where the point P is (1, $$-$$2, 3), then the area of the triangle PQR is equal to :

The acute angle between the planes P₁ and P₂, when P₁ and P₂ are the planes passing through the intersection of the planes $$5x + 8y + 13z - 29 = 0$$ and $$8x - 7y + z - 20 = 0$$ and the points (2, 1, 3) and (0, 1, 2), respectively, is :

Let the plane $$P:\overrightarrow r \,.\,\overrightarrow a = d$$ contain the line of intersection of two planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6$$ and $$\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7$$. If the plane P passes through the point $$\left( {2,3,{1 \over 2}} \right)$$, then the value of $${{|13\overrightarrow a {|^2}} \over {{d^2}}}$$ is equal to :

Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :

The shortest distance between the lines

$${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$ and $${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$, is :

If two straight lines whose direction cosines are given by the relations $$l + m - n = 0$$, $$3{l^2} + {m^2} + cnl = 0$$ are parallel, then the positive value of c is :

If the plane $$2x + y - 5z = 0$$ is rotated about its line of intersection with the plane $$3x - y + 4z - 7 = 0$$ by an angle of $${\pi \over 2}$$, then the plane after the rotation passes through the point :

If the lines $$\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$$ are co-planar, then the distance of the plane containing these two lines from the point ($$\alpha$$, 0, 0) is :

Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$, $$\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$ be three given vectors. Let $$\overrightarrow v $$ be a vector in the plane of $$\overrightarrow a $$ and $$\overrightarrow b $$ whose projection on $$\overrightarrow c $$ is $${2 \over {\sqrt 3 }}$$. If $$\overrightarrow v \,.\,\widehat j = 7$$, then $$\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$$ is equal to :

If the two lines $${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$$ and $${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$$ are perpendicular, then an angle between the lines l₂ and $${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$$ is :

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $$-$$ 3y + 5z = 8. If the mirror image of the point $$\left( {2, - {1 \over 2},2} \right)$$ in the rotated plane is B(a, b, c), then :

Let p be the plane passing through the intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$$ and $$\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$$, and the point (2, 1, $$-$$2). Let the position vectors of the points X and Y be $$\widehat i - 2\widehat j + 4\widehat k$$ and $$5\widehat i - \widehat j + 2\widehat k$$ respectively. Then the points :

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR² is equal to :

If the shortest distance between the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$$ and $${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is $${1 \over {\sqrt 3 }}$$, then the sum of all possible value of $$\lambda$$ is :

Let the points on the plane P be equidistant from the points ($$-$$4, 2, 1) and (2, $$-$$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :

Let a line $L_1$ pass through the origin and be perpendicular to the lines

$\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{i}+(2 \mathrm{t}-1) \hat{j}+(2 \mathrm{t}+4) \hat{k}$ and

$\mathrm{L}_3: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{i}+(3+2 \mathrm{~s}) \hat{j}+(2+\mathrm{s}) \hat{k}, \mathrm{t}, \mathrm{s} \in \mathbf{R}$.

If $(a, b, c), a \in \mathbf{Z}$, is the point on $\mathrm{L}_3$ at a distance of $\sqrt{17}$ from the point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$, then $(\mathrm{a}+\mathrm{b}+\mathrm{c})^2$ is equal to $\_\_\_\_$ .

Let the image of the point $\mathrm{P}(0,-5,0)$ in the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ be the point R and the image of the point $\mathrm{Q}\left(0, \frac{-1}{2}, 0\right)$ in the line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ be the point S . Then the square of the area of the parallelogram PQRS is $\_\_\_\_$ .

Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L intersect the $y z-$ plane at the point Q . Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the $y z$-plane at the point R . Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$ .

If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image

of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P , then $a+b$ is equal to $\_\_\_\_$ .

Let P be the image of the point $\mathrm{Q}(7,-2,5)$ in the line $\mathrm{L}: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $\mathrm{R}(5, \mathrm{p}, \mathrm{q})$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is _________.

Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26 \alpha(\mathrm{~PB})^2$ is _________ .

The square of the distance of the image of the point $$(6,1,5)$$ in the line $$\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$$, from the origin is __________.

Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(1,6,4)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$. Then $$2 \alpha+\beta+\gamma$$ is equal to ________

If the shortest distance between the lines $$\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$$ and $$\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$$ is $$\frac{44}{\sqrt{30}}$$, then the largest possible value of $$|\lambda|$$ is equal to _________.

Let $$P$$ be the point $$(10,-2,-1)$$ and $$Q$$ be the foot of the perpendicular drawn from the point $$R(1,7,6)$$ on the line passing through the points $$(2,-5,11)$$ and $$(-6,7,-5)$$. Then the length of the line segment $$P Q$$ is equal to _________.

Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.

Consider a line $$\mathrm{L}$$ passing through the points $$\mathrm{P}(1,2,1)$$ and $$\mathrm{Q}(2,1,-1)$$. If the mirror image of the point $$\mathrm{A}(2,2,2)$$ in the line $$\mathrm{L}$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+6 \gamma$$ is equal to __________.

A line passes through $$A(4,-6,-2)$$ and $$B(16,-2,4)$$. The point $$P(a, b, c)$$, where $$a, b, c$$ are non-negative integers, on the line $$A B$$ lies at a distance of 21 units, from the point $$A$$. The distance between the points $$P(a, b, c)$$ and $$Q(4,-12,3)$$ is equal to __________.

Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be the feet of perpendiculars from the point $$\mathrm{P}(a, a, a)$$ on the lines $$x=y, z=1$$ and $$x=-y, z=-1$$ respectively. If $$\angle \mathrm{QPR}$$ is a right angle, then $$12 a^2$$ is equal to _________.

Let a line passing through the point $$(-1,2,3)$$ intersect the lines $$L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$$ at $$M(\alpha, \beta, \gamma)$$ and $$L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$$ at $$N(a, b, c)$$. Then, the value of $$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$$ equals __________.

If $$\mathrm{d}_1$$ is the shortest distance between the lines $$x+1=2 y=-12 z, x=y+2=6 z-6$$ and $$\mathrm{d}_2$$ is the shortest distance between the lines $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$$, then the value of $$\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}$$ is :

Let O be the origin, and M and $$\mathrm{N}$$ be the points on the lines $$\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$$ and $$\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$$ respectively such that $$\mathrm{MN}$$ is the shortest distance between the given lines. Then $$\overrightarrow{O M} \cdot \overrightarrow{O N}$$ is equal to _________.

A line with direction ratios $$2,1,2$$ meets the lines $$x=y+2=z$$ and $$x+2=2 y=2 z$$ respectively at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. If the length of the perpendicular from the point $$(1,2,12)$$ to the line $$\mathrm{PQ}$$ is $$l$$, then $$l^2$$ is __________.

The lines $$\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$$ and $$\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$$ intersect at the point $$P$$. If the distance of $$\mathrm{P}$$ from the line $$\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$$ is $$l$$, then $$14 l^2$$ is equal to __________.

Let the image of the point $$\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$$ in the plane $$x-2 y+z-2=0$$ be P. If the distance of the point $$Q(6,-2, \alpha), \alpha > 0$$, from $$\mathrm{P}$$ is 13 , then $$\alpha$$ is equal to ___________.

Let the plane $$x+3 y-2 z+6=0$$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $$\mathrm{ABC}$$ is $$\left(\alpha, \beta, \frac{6}{7}\right)$$, then $$98(\alpha+\beta)^{2}$$ is equal to ___________.

Let the line $$l: x=\frac{1-y}{-2}=\frac{z-3}{\lambda}, \lambda \in \mathbb{R}$$ meet the plane $$P: x+2 y+3 z=4$$ at the point $$(\alpha, \beta, \gamma)$$. If the angle between the line $$l$$ and the plane $$P$$ is $$\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$$, then $$\alpha+2 \beta+6 \gamma$$ is equal to ___________.

Let a line $$l$$ pass through the origin and be perpendicular to the lines

$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and

$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.

If $$\mathrm{P}$$ is the point of intersection of $$l$$ and $$l_{1}$$, and $$\mathrm{Q}(\propto, \beta, \gamma)$$ is the foot of perpendicular from P on $$l_{2}$$, then $$9(\alpha+\beta+\gamma)$$ is equal to _____________.

Let the foot of perpendicular from the point $$\mathrm{A}(4,3,1)$$ on the plane $$\mathrm{P}: x-y+2 z+3=0$$ be N. If B$$(5, \alpha, \beta), \alpha, \beta \in \mathbb{Z}$$ is a point on plane P such that the area of the triangle ABN is $$3 \sqrt{2}$$, then $$\alpha^{2}+\beta^{2}+\alpha \beta$$ is equal to ___________.

Let $$\mathrm{P}_{1}$$ be the plane $$3 x-y-7 z=11$$ and $$\mathrm{P}_{2}$$ be the plane passing through the points $$(2,-1,0),(2,0,-1)$$, and $$(5,1,1)$$. If the foot of the perpendicular drawn from the point $$(7,4,-1)$$ on the line of intersection of the planes $$P_{1}$$ and $$P_{2}$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+\gamma$$ is equal to ___________.

Let $$\lambda_{1}, \lambda_{2}$$ be the values of $$\lambda$$ for which the points $$\left(\frac{5}{2}, 1, \lambda\right)$$ and $$(-2,0,1)$$ are at equal distance from the plane $$2 x+3 y-6 z+7=0$$. If $$\lambda_{1} > \lambda_{2}$$, then the distance of the point $$\left(\lambda_{1}-\lambda_{2}, \lambda_{2}, \lambda_{1}\right)$$ from the line $$\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$$ is ____________.

If the lines $$\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}$$ and $$\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}$$ intersect, then the magnitude of the minimum value of $$8 \alpha \beta$$ is _____________.

Let the image of the point $$\mathrm{P}(1,2,3)$$ in the plane $$2 x-y+z=9$$ be $$\mathrm{Q}$$. If the coordinates of the point $$\mathrm{R}$$ are $$(6,10,7)$$, then the square of the area of the triangle $$\mathrm{PQR}$$ is _____________.

The point of intersection $$\mathrm{C}$$ of the plane $$8 x+y+2 z=0$$ and the line joining the points $$\mathrm{A}(-3,-6,1)$$ and $$\mathrm{B}(2,4,-3)$$ divides the line segment $$\mathrm{AB}$$ internally in the ratio $$\mathrm{k}: 1$$. If $$\mathrm{a}, \mathrm{b}, \mathrm{c}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|$$ are coprime) are the direction ratios of the perpendicular from the point $$\mathrm{C}$$ on the line $$\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}$$, then $$|\mathrm{a}+\mathrm{b}+\mathrm{c}|$$ is equal to ___________.

Let $$\alpha x+\beta y+\gamma z=1$$ be the equation of a plane passing through the point $$(3,-2,5)$$ and perpendicular to the line joining the points $$(1,2,3)$$ and $$(-2,3,5)$$. Then the value of $$\alpha \beta y$$ is equal to _____________.

Let the line $$L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$$ intersect the plane $$2 x+y+3 z=16$$ at the point $$P$$. Let the point $$Q$$ be the foot of perpendicular from the point $$R(1,-1,-3)$$ on the line $$L$$. If $$\alpha$$ is the area of triangle $$P Q R$$, then $$\alpha^{2}$$ is equal to __________.

Let $$\theta$$ be the angle between the planes $$P_{1}: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$$ and $$P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15$$. Let $$\mathrm{L}$$ be the line that meets $$P_{2}$$ at the point $$(4,-2,5)$$ and makes an angle $$\theta$$ with the normal of $$P_{2}$$. If $$\alpha$$ is the angle between $$\mathrm{L}$$ and $$P_{2}$$, then $$\left(\tan ^{2} \theta\right)\left(\cot ^{2} \alpha\right)$$ is equal to ____________.

If the equation of the plane passing through the point $$(1,1,2)$$ and perpendicular to the line $$x-3 y+ 2 z-1=0=4 x-y+z$$ is $$\mathrm{A} x+\mathrm{B} y+\mathrm{C} z=1$$, then $$140(\mathrm{C}-\mathrm{B}+\mathrm{A})$$ is equal to ___________.

If $$\lambda_{1} < \lambda_{2}$$ are two values of $$\lambda$$ such that the angle between the planes $$P_{1}: \vec{r}(3 \hat{i}-5 \hat{j}+\hat{k})=7$$ and $$P_{2}: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \hat{k})=9$$ is $$\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$$, then the square of the length of perpendicular from the point $$\left(38 \lambda_{1}, 10 \lambda_{2}, 2\right)$$ to the plane $$P_{1}$$ is ______________.

Let the equation of the plane P containing the line $$x+10=\frac{8-y}{2}=z$$ be $$ax+by+3z=2(a+b)$$ and the distance of the plane $$P$$ from the point (1, 27, 7) be $$c$$. Then $$a^2+b^2+c^2$$ is equal to __________.

Let the co-ordinates of one vertex of $$\Delta ABC$$ be $$A(0,2,\alpha)$$ and the other two vertices lie on the line $${{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3}$$. For $$\alpha \in \mathbb{Z}$$, if the area of $$\Delta ABC$$ is 21 sq. units and the line segment $$BC$$ has length $$2\sqrt{21}$$ units, then $$\alpha^2$$ is equal to ___________.

If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {{z - 2} \over 0}$$ is $$\alpha$$, then 28$$\alpha^2$$ is equal to ____________.

Let the equation of the plane passing through the line $$x - 2y - z - 5 = 0 = x + y + 3z - 5$$ and parallel to the line $$x + y + 2z - 7 = 0 = 2x + 3y + z - 2$$ be $$ax + by + cz = 65$$. Then the distance of the point (a, b, c) from the plane $$2x + 2y - z + 16 = 0$$ is ____________.

If the shortest between the lines $${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$$ and $${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$$ is 6, then the square of sum of all possible values of $$\lambda$$ is :

The shortest distance between the lines $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$ and $${{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}$$ is equal to ________

Let a line with direction ratios $$a,-4 a,-7$$ be perpendicular to the lines with direction ratios $$3,-1,2 b$$ and $$b, a,-2$$. If the point of intersection of the line $$\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$$ and the plane $$x-y+z=0$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+\gamma$$ is equal to _________.

Let $$\mathrm{P}(-2,-1,1)$$ and $$\mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $$\alpha,-1, \beta$$, where both $$\alpha$$ and $$\beta$$ are integers of minimum absolute values, then $$\alpha^{2}+\beta^{2}$$ is equal to ____________.

Let the line $$\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$$ intersect the plane containing the lines $$\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$$ and $$4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \in \mathbb{R}$$ at the point $$P(\alpha, \beta, \gamma)$$. Then the value of $$\alpha+\beta+\gamma$$ equals _____________.

The largest value of $$a$$, for which the perpendicular distance of the plane containing the lines $$ \vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$$ and $$\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$$ from the point $$(2,1,4)$$ is $$\sqrt{3}$$, is _________.

The plane passing through the line $$L: l x-y+3(1-l) z=1, x+2 y-z=2$$ and perpendicular to the plane $$3 x+2 y+z=6$$ is $$3 x-8 y+7 z=4$$. If $$\theta$$ is the acute angle between the line $$L$$ and the $$y$$-axis, then $$415 \cos ^{2} \theta$$ is equal to _____________.

Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be two points on the line $$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$$ at a distance $$\sqrt{26}$$ from the point $$P(4,2,7)$$. Then the square of the area of the triangle $$P Q R$$ is ___________.

The line of shortest distance between the lines $$\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$$ and $$\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$$ makes an angle of $$\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$$ with the plane $$\mathrm{P}: \mathrm{a} x-y-z=0$$, $$(a>0)$$. If the image of the point $$(1,1,-5)$$ in the plane $$P$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta-\gamma$$ is equal to _________________.

Consider a triangle ABC whose vertices are A(0, $$\alpha$$, $$\alpha$$), B($$\alpha$$, 0, $$\alpha$$) and C($$\alpha$$, $$\alpha$$, 0), $$\alpha$$ > 0. Let D be a point moving on the line x + z $$-$$ 3 = 0 = y and G be the centroid of $$\Delta$$ABC. If the minimum length of GD is $$\sqrt {{{57} \over 2}} $$, then $$\alpha$$ is equal to ____________.

Let d be the distance between the foot of perpendiculars of the points P(1, 2, $$-$$1) and Q(2, $$-$$1, 3) on the plane $$-$$x + y + z = 1. Then d² is equal to ___________.

Let $${P_1}:\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$$ be a plane. Let P₂ be another plane which passes through the points (2, $$-$$3, 2), (2, $$-$$2, $$-$$3) and (1, $$-$$4, 2). If the direction ratios of the line of intersection of P₁ and P₂ be 16, $$\alpha$$, $$\beta$$, then the value of $$\alpha$$ + $$\beta$$ is equal to ________________.

Let the image of the point P(1, 2, 3) in the line $$L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ be Q. Let R ($$\alpha$$, $$\beta$$, $$\gamma$$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($$\alpha$$ + $$\beta$$ + $$\gamma$$) is equal to __________.

Let the mirror image of the point (a, b, c) with respect to the plane 3x $$-$$ 4y + 12z + 19 = 0 be (a $$-$$ 6, $$\beta$$, $$\gamma$$). If a + b + c = 5, then 7$$\beta$$ $$-$$ 9$$\gamma$$ is equal to ______________.

Let l₁ be the line in xy-plane with x and y intercepts $${1 \over 8}$$ and $${1 \over {4\sqrt 2 }}$$ respectively, and l₂ be the line in zx-plane with x and z intercepts $$ - {1 \over 8}$$ and $$ - {1 \over {6\sqrt 3 }}$$ respectively. If d is the shortest distance between the line l₁ and l₂, then d^$$-$$2 is equal to _______________.

Let the lines

$${L_1}:\overrightarrow r = \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right),\,\lambda \in R$$

$${L_2}:\overrightarrow r = \left( {\widehat i + 3\widehat j + \widehat k} \right) + \mu \left( {\widehat i + \widehat j + 5\widehat k} \right);\,\mu \in R$$,

intersect at the point S. If a plane ax + by $$-$$ z + d = 0 passes through S and is parallel to both the lines L₁ and L₂, then the value of a + b + d is equal to ____________.

Let a line having direction ratios, 1, $$-$$4, 2 intersect the lines $${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$$ and $${x \over 2} = {{y - 7} \over 3} = {z \over 1}$$ at the points A and B. Then (AB)² is equal to ___________.

If the shortest distance between the lines

$$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$$

and $$\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$$ is $$\sqrt {{2 \over 3}} $$, then the integral value of a is equal to ___________.