Three Dimensional Geometry · Mathematics · MHT CET

The co-ordinates of the point in which line joining $(1,1,1)$ and $(2,2,2)$ intersects the plane $x+y+\mathrm{z}=9$ are

The equation of plane passing through $(1,0,0)$ and $(0,1,0)$ and making an angle $45^{\circ}$ with the plane $x+y-3=0$ is

The distance of the point $(5,3,-1)$ from the plane passing through points $(2,1,0),(3,-2,4)$ and $(1,-3,3)$ is

The equation of a line passing through the point $(-1,2,3)$ and perpendicular to the lines $\frac{x}{2}=\frac{y-1}{-3}=\frac{z+2}{-2}$ and $\frac{x+3}{-1}=\frac{y+3}{2}=\frac{z-1}{3}$ is

A line L is passing through points $\mathrm{A}(1,3,2)$ and $\mathrm{B}(2,2,1)$. If mirror image of point $\mathrm{P}(1,1,-1)$ in the line L is $(x, y, z)$ then $x+y+\mathrm{z}=$

The lines $\frac{6 x-6}{18}=\frac{y+1}{3}=\frac{z-1}{5} \quad$ and $\frac{3 x+6}{12}=\frac{y-1}{3}=\frac{z+1}{2}$ are $\ldots$

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the XY plane and the YZ plane at points A and B respectively. The equation of line through the points A and B is

The distance of the point $\mathrm{A}(3,-4,5)$ from the plane $2 x+5 y-6 z=16$ measured along the line $\frac{x}{2}=\frac{y}{1}=\frac{z}{-2}$ is

The direction cosines of a normal to the plane passing through $(4,2,3),(-1,4,2)$ and $(3,2,1)$ are …..

The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2 y-z+1=0$ and $3 x-y-4 z+3=0$ is

If the foot of the perpendicular drawn from the origin to a plane is $\mathrm{P}(-1,-1,2)$, then equation of the plane is

The angle between lines whose direction cosines satisfy the equation $l+m+n=0$ and $l^2-\mathrm{m}^2-\mathrm{n}^2=0$, is

A triangle ABC is formed by $\mathrm{A}(1,-1,0)$, $B(3,5,3), C(-11,-5,6)$. The equation of internal angle bisector of angle $A$ is

The mirror image of the point $\mathrm{P}(-1,2,-4)$ in the plane $x-y-2 z+1=0$ is

If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the co-ordinate axes at points $A, B, C$ respectively, then area of the triangle ABC is

If the plane $\frac{x}{2}-\frac{y}{3}-\frac{\mathrm{z}}{5}=1$ cuts the co-ordinate axes in points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively, then the area of the triangle $A B C$ is

If the lines $x=a y-1=z-2$ and $x=3 y-2=\mathrm{bz}-2(\mathrm{ab} \neq 0)$ are coplanar, then

The Cartesian equation of the plane $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})$ is

If the line $\frac{x-3}{2}=\frac{y+5}{-1}=\frac{z+2}{2}$ lies in the plane $\alpha x+3 y-z+\beta=0$, then values of $\alpha$ and $\beta$ respectively are ….

The lines $\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})$ and $\overline{\mathrm{r}}=(4 \hat{\mathrm{i}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})$ are

The lines $\frac{x-0}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda}$ and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{\lambda}$ are coplanar and $p$ is the plane containing these lines, then which of following point does not lie on the plane.

The length of the foot of the perpendicular from the point $\left(1, \frac{3}{2}, 2\right)$ to the plane $2 x-2 y+4 z+17=0$ is

If the lines $\frac{1-x}{2}=\frac{7 y+4}{2 \lambda}=\frac{2 z-5}{2}$ and $\frac{7-7 x}{3 \lambda}=\frac{y-1}{7}=\frac{6-\mathrm{z}}{5}$ are at right angle, then the value of $\lambda$ is

Let M and N be foots of the perpendiculars drawn from the point $\mathrm{P}(\mathrm{a}, \mathrm{a}, \mathrm{a})$ on the lines $x-y=0, \mathrm{z}=1$ and $x+y=0, \mathrm{z}=-1$ respectively and if $\angle \mathrm{MPN}=90^{\circ}$ then $\mathrm{a}^2=$

If $\theta$ is the angle between the lines whose direction cosines are given by $6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{~m}=0$ and $3 l+\mathrm{m}+5 \mathrm{n}=0$, then $\sin \theta=$

The angle between the lines $x=y, z=0$ and $y=0, \mathrm{z}=0$ is

If the foot of the perpendicular drawn from the origin to a plane is $\mathrm{P}(2,-1,4)$, then the equation of the plane is

Let the plane passing through point $(2,1,-1)$ containing line joining the points $(1,3,2)$ and $(1,2,1)$ makes intercepts $\mathrm{p}, \mathrm{q}, \mathrm{r}$ on co-ordinate axes, then $\mathrm{p}+\mathrm{q}+\mathrm{r}=$

The angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=6$ is $\cos ^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is

The equation of the line passing through the point of intersection of $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and also through the point ( $2,1,-2$ ) is

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then the value of k is

The distance of the point $\mathrm{P}(3,4,4)$ from the point of intersection of the line joining the points $\mathrm{Q}(3,-4,-5), \mathrm{R}(2,-3,1)$ and the plane $2 x+y+z=7$ is

The equation of the plane containing the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to the plane containing the lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{1}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{1}$ is

The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

The line L is passing through $(1,2,3)$. The distance of any point on the line L from the line $\overline{\mathrm{r}}=(3 \lambda-1) \hat{\mathrm{i}}+(-2 \lambda+3) \hat{\mathrm{j}}+(4+\lambda) \hat{\mathrm{k}}$ is constant. Then the line L does not pass through the point

The distance of the plane $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ from the origin is

If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, \mathrm{z}=1$, then $\alpha+\beta=$

The angle between the lines $x-3 y-4=0,4 y-z+5=0$ and $x+3 y-11=0,2 y-z+6=0$ is

If the planes $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\lambda \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$ and $\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\mu \hat{\mathrm{k}})=5$ are parallel, then $\lambda+\mu=$

The perimeter of a square whose two sides have equations $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ is

If the sum of the squares of the distances of a point $\mathrm{P}(x, y, z)$ from the three co-ordinate axes is 324 , then the distance of point P from the origin is ….

The angle between the lines $3 x=2 y=-\mathrm{z}$ and $-x=6 y=-4 z$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{\mathrm{k}}=\frac{\mathrm{z}}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{\mathrm{z}}{\mathrm{k}}$ are coplanar, then the equation of the plane containing these lines are

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3 y-\alpha z+\beta=0$, then the value of $(\beta-\alpha)$ is equal to

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3 y-6=0$, are

The distance between the line $\overline{\mathrm{r}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and the plane $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=4$ is

The angle between the lines whose direction cosines are $\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}$ and $\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}$ is

The length of the altitude through the point $D$ of tetrahedron where the vertices of the tetrahedron are $A(2,3,1), B(4,1,-2), C(6,3,7), D(-5,-4,8)$, is

The angle between the lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{\mathrm{~m}}=\frac{y-3}{\mathrm{n}}=\frac{\mathrm{z}-1}{l}$, where $l>\mathrm{m}>\mathrm{n}$ and $1, \mathrm{~m}, \mathrm{n}$ are roots of the equation $x^3+x^2-4 x-4=0$, is

The distance of the point $\mathrm{P}(3,8,2)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3 x+2 y-2 z+15=0$ is

The shortest distance between the lines $\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})$ is

If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha>0$ is 5 units, then the foot of the perpendicular from P to the plane is

The equation of the plane containing the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-4}{-2}$ and the point $(0,5,0)$ is

The equation of the plane passing through the point of intersection of the planes $2 x-y+z-3=0$ and $4 x-3 y+5 z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$ Then $\alpha+\beta+\gamma+d=$

The distance of the point $(2,4,0)$ from the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ is

The co-ordinates of the point where the line joining the points $(2,-3,1)$ and $(3,-4,-5)$ and intersects the plane $2 x+y+z=7$ are

If the line $\frac{x+1}{3}=\frac{y-k}{7}=\frac{z-4}{8}$ lies in the plane $2 x+\mathrm{p} y+7 z-41=0$ which is perpendicular to the plane $x+4 y-2 z+13=0$ then $\mathrm{k}=$

The direction cosines of the line $x-y+2 z=5$ and $3 x+y+z=6$ are

If the angle between the planes $x-2 y+3 z-5=0$ and $x+\alpha y+2 z+7=0$ is $\cos ^{-1}\left(\frac{1}{14}\right)$ then the difference between the values of $\alpha$ is

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

The acute angle between the lines $x=-2+2 \mathrm{t}, y=3-4 \mathrm{t}, \mathrm{z}=-4+\mathrm{t}$ and $x=-2-\mathrm{t}, y=3+2 \mathrm{t}, \mathrm{z}=-4+3 \mathrm{t}$ is

If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the co-ordinate axes at points $\mathrm{A}, \mathrm{B}$ and C , then the area of the triangle ABC is

A plane passes through $(2,1,2)$ and $(1,2,1)$ and parallel to the line $2 x=3 y$ and $\mathrm{z}=1$, then the plane also passes through the point

The coordinates of the foot of the perpendicular drawn from a point $\mathrm{P}(-1,1,2)$ to the plane $2 x-3 y+z-11=0$ are

The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{\mathrm{k}}=\frac{y-3}{1}=\frac{\mathrm{z}-3}{2}$ are coplanar, hence $\mathrm{k}=$

If the shortest distance between the lines $\bar{r}_1=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in \mathbb{R}, \alpha>0 \quad$ and $\bar{r}_2=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$, is 9 , then the value of $\alpha$ is

The co-ordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ is

A line having direction ratios $1,-4,2$ intersects the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ resp., then co-ordinates of points A and B are

A plane makes positive intercepts of unit length on each of $X$ and $Y$ axis. If it passes through the point $(-1,1,2)$ and makes angle $\theta$ with the X -axis, then $\theta$ is

The equation of plane through the point $(2,-1,-3)$ and parallel to lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

The equation of the plane, passing through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $Y$-axis is

A line with positive direction cosines passes through the point $\mathrm{P}(2,-1,2)$ and makes equal angles with co-ordinate axes. The line meets the plane $2 x+y+z=9$ at point Q. Then the length of the line segment PQ equals

If the distance between the plane Ax-2y+z $=\mathrm{d}$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$ units, then $|d|$ is

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is

A line makes $45^{\circ}$ angle with positive X -axis and makes equal angles with positive Y -axis ad Z-axis respectively, then the sum of the three angles which the line makes with positive X -axis, Y -axis and Z -axis is

If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then k has the value

The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y+4 z=5$, which is perpendicular to the plane $x-y+z=0$, is

The equation of a line passing through the point $(2,-1,1)$ and parallel to the line joining the points $\hat{i}+2 \hat{j}+2 \hat{k}$ and $-\hat{i}+4 \hat{j}+\hat{k}$ is

The foot of the perpendicular drawn from origin to a plane is $\mathrm{M}(2,1,-2)$, then vector equation of the plane is

Let $\mathrm{L}_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{\mathrm{z}-6}{1}$ and $\mathrm{L}_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is

The perpendicular distance from the origin to the plane containing the two lines $\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$, is

Let $P(2,1,5)$ be a point in space and $Q$ be a point on the line $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(-3 \hat{i}+\hat{j}+5 \hat{k})$. Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $3 x-y+4 z=1$ is

The centroid of tetrahedron with vertices $\mathrm{P}(5,-7,0), \mathrm{Q}(\mathrm{a}, 5,3), \mathrm{R}(4,-6, b)$ and $\mathrm{S}(6, \mathrm{c}, 2)$ is $(4,-3,2)$, then the value of $2 a+3 b+c$ is equal to

A variable plane passes through the fixed point $(3,2,1)$ and meets $X, Y$ and $Z$ axes at points $A$, B and C respectively. A plane is drawn parallel to YZ - plane through A , a second plane is drawn parallel to ZX -plan through B , a third plane is drawn parallel to XY - plane through C . Then locus of the point of intersection of these three planes, is

The distance of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along the line $x=y=\mathrm{z}$ is __________ units.

If for some $\alpha \in \mathbb{R}$, the lines $\mathrm{L}_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\mathrm{L}_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar, then the line $L_2$ passes through the point

Let $P(3,2,6)$ be a point in space and $Q$ be a point on the line $\bar{r}=\hat{i}-\hat{j}+2 \hat{k}+\mu(-3 \hat{i}+\hat{j}+5 \hat{k})$. Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $x-4 y+3 z=1$ is

The perpendicular distance of the origin from the plane $2 x+y-2 z-18=0$ is

The plane $2 x+3 y+4 z=1$ meets $X$-axis in $A$, Y -axis in B and Z -axis in C . Then the centroid of $\triangle A B C$ is

If the lines $\frac{x+1}{-10}=\frac{y+k}{-1}=\frac{z-4}{1} \quad$ and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$ intersect each other, then the value of $k$ is

The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

The area of the triangle with vertices $(1,2,0)$, $(1,0,2)$ and $(0,3,1)$ is

If the volume of tetrahedron whose vertices are $A \equiv(1,-6,10), B \equiv(-1,-3,7), C \equiv(5,-1, k)$ and $D \equiv(7,-4,7)$ is 11 cu . units, then the value of $k$ is

The vector equation of the plane passing through the point $\mathrm{A}(1,2,-1)$ and parallel to the vectors $2 \hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+3 \hat{k}$ is

The shortest distance between lines $\bar{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=(2 \hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then the value of k is

The projection of $\overline{\mathrm{AB}}$ on $\overline{\mathrm{CD}}$, where $A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$ and $\mathrm{D} \equiv(7,0,10)$ is

The equation of the plane through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

Equation of the plane, through the points $(-1,2,-2)$ and $(-1,3,2)$ and perpendicular to $y \mathrm{z}$ - plane, is

If the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{-1}=\frac{y-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then k is equal to

If the line, $\frac{x-3}{2}=\frac{y+2}{1}=\frac{z+4}{3}$ lies in the plane, $\ell x+m y-z=9$, then $\ell^2+m^2$ is equal to

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$, then $(\alpha, \beta)=$

A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,2,1)$. The distance of the plane from the point $(2,3,4)$ is

The value of m such that $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ lies in the plane $2 x-4 y+z=7$ is

A line with positive direction cosines passes through the point $\mathrm{P}(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x+y+\mathrm{z}=9$ at point Q . The length of the line segment PQ equals $\qquad$ units.

Let L be the line of intersection of the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$. If L makes an angle $\alpha$ with the positive X -axis, then $\cos \alpha$ equals

The equation of the plane, passing through the point $(1,1,1)$ and perpendicular to the planes $2 x+y-2 z=5$ and $3 x-6 y-2 z=7$, is

The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ is

The value of m , such that $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-m}{2}$ lies in the plane $2 x-4 y+z=7$, is

The length of the perpendicular from the point $\mathrm{A}(1,-2,-3)$ on the line $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}$ is

If the points $(1,-1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane $3 x-4 y-12 z+13=0$, then the sum of all possible values of $\lambda$ is

Let P be a plane passing through the points $(2,1,0),(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$. Then image of $R$ in the plane $P$ is

The equation of the plane, passing through the point $(-1,2,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$, is

The co-ordinates of the point where the line through $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the $x y$-plane are

The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} \quad$ and $\frac{x-1}{\mathrm{k}}=\frac{y-4}{2}=\frac{\mathrm{z}-5}{1}$ are coplanar if

Let $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ be two given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

Let $a, b \in R$. If the mirror image of the point $\mathrm{p}(\mathrm{a}, 6,9)$ w.r.t. line $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20, b,-a-9)$, then $|a+b|$ is equal to

A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,-2,1)$. The distance of the plane from the point $(1,2,2)$ is

Let $\mathrm{L}_1$ $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2 x-y-2 z=5$ and $3 x-6 y+2 z=7$ is

Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

If $A(-4,5, P), B(3,1,4)$ and $C(-2,0, q)$ are the vertices of a triangle $A B C$ and $G(r, q, 1)$ is its centroid, then the value of $2 p+q-r$ is equal to

On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ ?

Equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is

The value of $m$, such that $\frac{x-4}{1}=\frac{y-2}{1}=\frac{2 z-m}{3}$ lies in the plane $2 x-5 y+2 z=7$, is

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2 x-y+z+3=0$ is the line

Let $\mathrm{P}(2,3,6)$ be a point in space and Q be a point on the line $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(-3 \hat{i}+\hat{j}+5 \hat{k})$. Then the value of $\mu$ for which vector $\overline{\mathrm{PQ}}$ is parallel to the plane $x-4 y+4 z=1$ is

Distance between the parallel lines $\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}$ and $\frac{x+4}{3}=\frac{y-3}{-2}=\frac{z+2}{1}$ is

The equation of the plane, passing through the mid point of the line segment of join of the points $\mathrm{P}(1,2,5)$ and $\mathrm{Q}(3,4,3)$ and perpendicular to it, is

The area of the triangle, whose vertices are $A \equiv(1,-1,2), B \equiv(2,1,-1)$ and $C \equiv(3,-1,2)$, is

The equation of the line, through $\mathrm{A}(1,2,3)$ and perpendicular to the vector $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\hat{i}+3 \hat{j}+2 \hat{k}$, is

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

The incentre of the triangle whose vertices are $P(0,3,0), Q(0,0,4)$ and $R(0,3,4)$ is

The vector equation of a line whose Cartesian equations are $y=2,4 x-3 z+5=0$ is

The Cartesian equation of the plane, passing through the points $(3,1,1),(1,2,3)$ and $(-1,4,2)$, is

The equation of the line passing through the point $(-1,3,-2)$ and perpendicular to each of the lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$, is

If the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $\ell x+m y-z=9$, then $\ell^2+m^2$ is

The mirror image of $$\mathrm{P}(2,4,-1)$$ in the plane $$x-y+2 z-2=0$$ is $$(\mathrm{a}, \mathrm{b}, \mathrm{c})$$, then the value of $$a+b+c$$ is

If the lines $$\frac{x-\mathrm{k}}{2}=\frac{y+1}{3}=\frac{\mathrm{z}-1}{4}$$ and $$\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{\mathrm{z}}{1}$$ intersect, then the value of $$\mathrm{k}$$ is

A vector parallel to the line of intersection of the planes $$\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$$ and $$\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$$ is

The length of the perpendicular drawn from the point $$(1,2,3)$$ to the line $$\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$$ is

If $$\triangle \mathrm{ABC}$$ is right angled at $$\mathrm{A}$$, where $$A \equiv(4,2, x), \mathrm{B} \equiv(3,1,8)$$ and $$C \equiv(2,-1,2)$$, then the value of $$x$$ is

The angle between the lines, whose direction cosines $$l, \mathrm{~m}, \mathrm{n}$$ satisfy the equations $$l+\mathrm{m}+\mathrm{n}=0$$ and $$2 l^2+2 \mathrm{~m}^2-\mathrm{n}^2=0$$, is

Equation of the plane passing through $$(1,-1,2)$$ and perpendicular to the planes $$x+2 y-2 z=4$$ and $$3 x+2 y+z=6$$ is

A line with positive direction cosines passes through the point $$\mathrm{P}(2,-1,2)$$ and makes equal angles with the co-ordinate axes. The line meets the plane $$2 x+y+z=9$$ at point $$\mathrm{Q}$$. The length of the line segment $$P Q$$ equals

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

Consider the lines $$\mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2}$$

$$\mathrm{L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{\mathrm{z}-3}{3}$$, then the unit vector perpendicular to both $$\mathrm{L}_1$$ and $$\mathrm{L}_2$$ is

A tetrahedron has vertices at $$P(2,1,3), Q(-1,1,2), R(1,2,1)$$ and $$O(0,0,0)$$, then angle between the faces $$O P Q$$ and $$P Q R$$ is

A plane is parallel to two lines whose direction ratios are $$2,0,-2$$ and $$-2,2,0$$ and it contains the point $$(2,2,2)$$. If it cuts coordinate axes at $$A, B, C$$, then the volume of the tetrahedron $$O A B C$$ (in cubic units) is

The incentre of the $$\triangle A B C$$, whose vertices are $$A(0,2,1), B(-2,0,0)$$ and $$C(-2,0,2)$$, is

The acute angle between the line joining the points $$(2,1,-3),(-3,1,7)$$ and a line parallel to $$\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$$ through the point $$(-1,0,4)$$ is

The foot of the perpendicular from the point $$(1,2,3)$$ on the line $$\mathbf{r}=(6 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$$ has the coordinates

The distance of the point $$(1,6,2)$$ from the point of intersection of the line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$$ and the plane $$x-y+z=16$$ is

A line drawn from the point $$\mathrm{A}(1,3,2)$$ parallel to the line $$\frac{x}{2}=\frac{y}{4}=\frac{z}{1}$$, intersects the plane $$3 x+y+2 z=5$$ in point $$\mathrm{B}$$, then co-ordinates of point $$\mathrm{B}$$ are

A line $$\mathrm{L}_1$$ passes through the point, whose p. v. (position vector) $$3 \hat{i}$$, is parallel to the vector $$-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$$. Another line $$\mathrm{L}_2$$ passes through the point having p.v. $$\hat{i}+\hat{j}$$ is parallel to vector $$\hat{i}+\hat{k}$$, then the point of intersection of lines $$L_1$$ and $$L_2$$ has p.v.

The equation of the line passing through the point $$(-1,3,-2)$$ and perpendicular to each of the lines $$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$$ and $$\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$$ is

If $$A(1,4,2)$$ and $$C(5,-7,1)$$ are two vertices of triangle $$A B C$$ and $$G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$$ is centroid of the triangle $$A B C$$, then the mid point of side $$B C$$ is

The distance of the point $$(-1,-5,-10)$$ from the point of intersection of the line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$$ and the plane $$x-y+z=5$$ is

The equation of the line, passing through $$(1,2,3)$$ and parallel to planes $$x-y+2 z=5$$ and $$3 x+y+z=6$$, is

The shortest distance (in units) between the lines $$\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$$ and $$\bar{r}=(2 \hat{i}-2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+2 \hat{j})$$ is

The length (in units) of the projection of the line segment, joining the points $$(5,-1,4)$$ and $$(4,-1,3)$$, on the plane $$x+y+z=7$$ is

If the volume of tetrahedron, whose vertices are $$\mathrm{A}(1,2,3), \mathrm{B}(-3,-1,1), \mathrm{C}(2,1,3)$$ and $$D(-1,2, x)$$ is $$\frac{11}{6}$$ cubic units, then the value of $$x$$ is

Equation of plane containing the line $$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$$ and perpendicular to the plane containing the lines $$\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$$ and $$\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$$ is

The centroid of tetrahedron with vertices at $$\mathrm{A}(-1,2,3), \mathrm{B}(3,-2,1), \mathrm{C}(2,1,3)$$ and $$\mathrm{D}(-1,-2,4)$$ is

A plane is parallel to two lines whose direction ratios are $$1,0,-1$$ and $$-1,1,0$$ and it contains the point $$(1,1,1)$$. If it cuts the co-ordinate axes at $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$, then the volume of the tetrahedron $$\mathrm{OABC}$$ (in cubic units) is

The equation of the plane through $$(-1,1,2)$$ whose normal makes equal acute angles with co-ordinate axes is

The distance of the point $$\mathrm{P}(-2,4,-5)$$ from the line $$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$ is

If the line $$\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$$ and $$\frac{7-7 x}{3 \mathrm{p}}=\frac{y-5}{1}=\frac{6-\mathrm{z}}{5}$$ are at right angles, then $$\mathrm{p}=$$

If the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$$ meets the plane $$x+2 y+3 z=15$$ at the point $$P$$, then the distance of $$\mathrm{P}$$ from the origin is

The equation of line passing through the point $$(1,2,3)$$ and perpendicular to the lines $$\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$$ and $$\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$$ is

The angle between the line $$\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$$ and plane $$x-2 y-\lambda z=3$$ is $$\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$$, then value of $$\lambda$$ is

If the direction cosines $$l, \mathrm{~m}, \mathrm{n}$$ of two lines are connected by relations $$l-5 \mathrm{~m}+3 \mathrm{n}=0$$ and $$7 l^2+5 \mathrm{~m}^2-3 \mathrm{n}^2=0$$, then value of $$l+\mathrm{m}+\mathrm{n}$$ is

The mirror image of the point $$(1,2,3)$$ in a plane is $$\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$$. Thus, the point _________ lies on this plane.

A plane is parallel to two lines, whose direction ratios are $$1,0,-1$$ and $$-1,1,0$$ and it contains the point $$(1,1,1)$$. If it cuts co-ordinate axes $$(\mathrm{X}, \mathrm{Y}, \mathrm{Z}$$ - axes resp.) at $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$, then the volume of the tetrahedron $$\mathrm{OABC}$$ is _________ cu. units.

The lines $$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \quad$$ and $$\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$$

The vector equation of the line $$2 x+4=3 y+1=6 z-3$$ is

The plane through the intersection of planes $$x+y+z=1$$ and $$2 x+3 y-z+4=0$$ and parallel to $$\mathrm{Y}$$-axis also passes through the point

The perpendicular distance of the origin from the plane $$x-3 y+4 z-6=0$$ is

Two lines $$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$$ and $$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4} \quad$$ intersect at the point R. Then reflection of $$\mathrm{R}$$ in the $$x y$$-plane has co-ordinates

The line $$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$$ lies in the plane $$x+3 y-\alpha z+\beta=0$$, then the value of $$\alpha^2+\alpha \beta+\beta^2$$ is

Let $$\mathrm{P}$$ be a plane passing through the points $$(2,1,0),(4,1,1)$$ and $$(5,0,1)$$ and $$R$$ be the point $$(2,1,6)$$. Then image of $$R$$ in the plane $$P$$ is

The co-ordinates of the point, where the line through $$A(3,4,1)$$ and $$B(5,1,6)$$ crosses the $$\mathrm{XZ}$$-plane, are

$$\mathrm{ABC}$$ is a triangle in a plane with vertices $$\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$$ and $$\mathrm{C}(\lambda, 5, \mu)$$. If median through $$\mathrm{A}$$ is equally inclined to the co-ordinate axes, then value of $$\lambda+\mu$$ is

If a line $$\mathrm{L}$$ is the line of intersection of the planes $$2 x+3 y+z=1$$ and $$x+3 y+2 z=2$$. If line $$\mathrm{L}$$ makes an angle $$\alpha$$ with the positive $$\mathrm{X}$$-axis, then the value of $$\sec \alpha$$ is

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

The co-ordinates of the point, where the line $$\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$$ meets the plane $$2 x+4 y-\mathrm{z}=3$$, are

The equation of a plane, containing the line of intersection of the planes $$2 x-y-4=0$$ and $$y+2 z-4=0$$ and passing through the point $$(2,1,0)$$, is

The foot of the perpendicular drawn from the origin to the plane is $$(4,-2,5)$$, then the Cartesian equation of the plane is

A vector $$\overrightarrow{\mathrm{n}}$$ is inclined to $$\mathrm{X}$$-axis at $$45^{\circ}$$, $$\mathrm{Y}$$-axis at $$60^{\circ}$$ and at an acute angle to Z-axis If $$\overrightarrow{\mathrm{n}}$$ is normal to a plane passing through the point $$(-\sqrt{2}, 1,1)$$, then equation of the plane is

If the Cartesian equation of a line is $$6 x-2=3 y+1=2 z-2$$, then the vector equation of the line is

The distance between parallel lines

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

$$\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$$ is :

A line makes the same angle '$$\alpha$$' with each of the $$x$$ and $$y$$ axes. If the angle '$$\theta$$', which it makes with the $$z$$-axis, is such that $$\sin ^2 \theta=2 \sin ^2 \alpha$$, then the angle $$\alpha$$ is

A tetrahedron has verticles $$P(1,2,1), Q(2,1,3), R(-1,1,2)$$ and $$O(0,0,0)$$. Then the angle between the faces $$O P Q$$ and $$P Q R$$ is

The Cartesian equation of a line passing through $$(1,2,3)$$ and parallel to $$x-y+2 z=5$$ and $$3 x+y+z=6$$ is

The equation of the plane passing through the points $$(2,3,1),(4,-5,3)$$ and parallel to $$X$$-axis is

The equation of the plane which passes through (2, $$-$$3, 1) and is normal to the line joining the points (3, 4, $$-$$1) and (2, $$-$$1, 5) is given by

If $$G(3,-5, r)$$ is the centroid of $$\triangle A B C$$, where $$A \equiv(7,-8,1), B \equiv(p, q, 5), C \equiv(q+1,5 p, 0)$$ are vertices of the triangle $$A B C$$, then the values of $$p, q, r$$ are respectively

If the lines $$\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$$ and $$\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}$$ are perpendicular to each other, then $$\lambda=$$

The co-ordinates of the points on the line $$\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$$ at a distance of 12 units from the point A($$-$$2, 1, $$-$$1) are

If the vector equation of the plane $$\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$$ in scalar product form is given by $$\overline{\mathrm{r}} \cdot(3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=\alpha$$ then $$\alpha=$$

If the lines $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$$ and $$\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}$$ intersect each other, then value of m is

The length of perpendicular drawn from the point $$2 \hat{i}-\hat{j}+5 \hat{k}$$ to the line $$\overline{\mathrm{r}}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})$$ is

Equation of the plane passing through the point $$(1,2,3)$$ and parallel to the plane $$2 x+3 y-4 z=0 $$

If $$\mathrm{A}$$ and $$\mathrm{B}$$ are the foot of the perpendicular drawn from the point $$\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$$ to the planes $$\mathrm{YZ}$$ and $$\mathrm{ZX}$$ respectively, then the equation of the plane through the points $$\mathrm{A}, \mathrm{B}$$, and $$\mathrm{O}$$ is (where $$\mathrm{O}$$ is the origin)

If $$\mathrm{A}=(-2,2,3), \mathrm{B}=(3,2,2), \mathrm{C}=(4,-3,5)$$ and $$\mathrm{D}=(7,-5,-1)$$ Then the projection of $$\overline{\mathrm{AB}}$$ on $$\overline{\mathrm{CD}}$$ is

The Cartesian equation of a plane which passes through the points $$\mathrm{A}(2,2,2)$$ and making equal nonzero intercepts on the co-ordinate axes is

The co-ordinates of the foot of the perpendicular drawn from the point $$2 \hat{i}-\hat{j}+5 \hat{k}$$ to the line $$\vec{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})$$ are

If A(3, 2, $$-$$1), B($$-$$2, 2, $$-$$3) and D($$-$$2, 5, $$-$$4) are the vertices of a parallelogram, then the area of the parallelogram is

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

The coordinates of the foot of the perpendicular drawn from the origin to the plane $$2 x+y-2 z=18$$ are

The vector equation of the line passing through $$\mathrm{P}(1,2,3)$$ and $$\mathrm{Q}(2,3,4)$$ is

Equation of planes parallel to the plane $$x-2y+2z+4=0$$ which are at a distance of one unit from the point (1, 2, 3) are

The area of triangle with vertices $$(1,2,0),(1,0, a)$$ and $$(0,3,1)$$ is $$\sqrt{6}$$ sq. units, then the values of '$$a$$' are

If $$\mathrm{G}(4,3,3)$$ is the centroid of the triangle $$\mathrm{ABC}$$ whose vertices are $$\mathrm{A}(\mathrm{a}, 3,1), \mathrm{B}(4,5, \mathrm{~b})$$ and $$C(6, c, 5)$$, then the values of $$a, b, c$$ are

The d.r.s. of the normal to the plane passing through the origin and the line of intersection of the planes $$x+2 y+3 z=4$$ and $$4 x+3 y+2 z=1$$ are

The line $$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$$ lies in the plane $$x+3 y-\alpha z+\beta=0$$, then value of $$\alpha \beta$$ is

If the points $$P(4,5, x), Q(3, y, 4)$$ and $$R(5,8,0)$$ are collinear, then the value of $$x+y$$ is

A line drawn from a point $$A(-2,-2,3)$$ and parallel to the line $$\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$$ meets the $$\mathrm{YOZ}$$ plane in point $$\mathrm{P}$$, then the co-ordinates of the point $$\mathrm{P}$$ are

The Cartesian equation of a line is $$3 x+1=6 y-2=1-z$$, then its vector equation is

The plane $$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$$ cuts the $$X$$-axis at A, Y-axis at B and Z-axis at C, then the area of $$\triangle \mathrm{ABC}=$$

If a plane meets the axes $$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$$ in $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ respectively such that centroid of $$\triangle \mathrm{ABC}$$ is $$(1,2,3)$$, then the equation of the plane is

The shortest distance between lines $$\bar{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$$ and $$\bar{r}=(\hat{r}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$$ is

The direction cosines $$\ell, \mathrm{m}, \mathrm{n}$$ of the line $$\frac{\mathrm{x}+2}{2}=\frac{2 \mathrm{y}-5}{3} ; \mathrm{z}=-1$$ are

Equation of the plane passing through the point (2, 0, 5) and parallel to the vectors $$\widehat i - \widehat j + \widehat k$$ and $$3\widehat i + 2\widehat j - \widehat k$$ is

The co-ordinates of the point $$\mathrm{P} \equiv(1,2,3)$$ and $$\mathrm{O} \equiv(0,0,0)$$, then the direction cosines of $$\overline{\mathrm{OP}}$$ are

The equation of the plane containing the line $$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$$ and the point $$(0,7,-7)$$ is

The equation of a line passing through $$(3,-1,2)$$ and perpendicular to the lines $$\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$$ and $$\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})$$ is

The area of the parallelogram with vertices A(1, 2, 3), B(1, 3, a), C(3, 8, 6) and D(3, 7, 3) is $$\sqrt{265}$$ sq. units, then a =

If the lines $\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles, then $\lambda=$

The Cartesian equation of the plane passing through the point A(7, 8, 6) and parallel to the XY plane is

The equation of the plane passing through $$(-2,2,2)$$ and $$(2,-2,-2)$$ and perpendicular to the plane $$9 x-13 y-3 z=0$$ is

The Cartesian equation of the line passing through the points A(2, 2, 1) and B(1, 3, 0) is

The Cartesian equation of the plane $$\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$$ is

The equation of the plane that contains the line of intersection of the planes. $$x+2 y+3 z-4=0$$ and $$2 x+y-z+5=0$$ and is perpendicular to the plane $$5 x+3 y-6 z+8=0$$ is

The vector equation of the line whose Cartesian equations are y = 2 and 4x $$-$$ 3z + 5 = 0 is

The Cartesian equation of the plane passing through the point $$(0,7,-7)$$ and containing the line $$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$$ is

If the lines $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$$ and $$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$$ intersect, then the values of $$k$$ is

The parametric equations of a line passing through the points $$\mathrm{A}(3,4,-7)$$ and $$\mathrm{B}(1,-1,6)$$ are

The angle between a line with direction ratios 2, 2, 1 and a line joining (3, 1, 4) and (7, 2, 12) is

If the line $$\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}$$ lies in the plane $$3 x-14 y+6 z+49=0$$, then the value of $$m$$ is

The point $P$ lies on the line $A, B$ where $A=(2,4,5)$ and $B \equiv(1,2,3)$. If $z$ co-ordinate of point $P$ is 3 , the its $y$ co-ordinate is

A line makes angles $\alpha, \beta, \gamma$ with the co-ordinate axes and $\alpha+\beta=90^{\circ}$, then $\gamma=$

The equations of planes parallel to the plane $x+2 y+2 z+8=0$, which are at a distance of 2 units from the point $(1,1,2)$ are

The equation of a plane containing the point $(1,-1,2)$ and perpendicular to the planes $2 x+3 y-2 z=5$ and $x+2 y-3 z=8$ is

The equation of the line passing through $(1,2,3)$ and perpendicular to the lines $x-1=\frac{y+2}{2}=\frac{z+4}{4}$ and $\frac{x-1}{2}=\frac{y-2}{2}=z+3$ is

If the plane $$2 x+3 y+5 z=1$$ intersects the co-ordinate axes at the points $$A, B, C$$, then the centroid of $$\triangle A B C$$ is

The direction co-sines of the line which bisects the angle between positive direction of $$Y$$ and $$Z$$ axes are

The angle between the lines $$\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$$ and $$\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$$ is

If the line $$r=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$$ is parallel to the plane $$r \cdot(3 \hat{i}-2 \hat{\mathbf{j}}+m \hat{\mathbf{k}})=10$$, then the value of $$m$$ is

The points $$A(-a,-b), B(0,0), C(a, b)$$ and $$D\left(a^2, a b\right)$$ are

The cosine of the angle included between the lines $$\mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$$ and $$\mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\mu(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}})$$ where $$\lambda, \mu \in R$$ is.

If the foot of perpendicular drawn from the origin to the plane is $$(3,2,1)$$, then the equation of plane is

The angle between the line $$r =(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+\hat{\mathbf{j}})$$ and the plane $$\mathbf{r} \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=8$$ is

The direction cosines of a line which is perpendicular to lines whose direction ratios are $$3,-2,4$$ and $$1,3,-2$$ are

If the lines given by $$\frac{x-1}{2 \lambda}=\frac{y-1}{-5}=\frac{z-1}{2}$$ and $$\frac{x+2}{\lambda}=\frac{y+3}{\lambda}=\frac{z+5}{1}$$ are parallel, then the value of $$\lambda$$ is

The vector equation of the plane $\mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ in scalar product form is $\mathbf{r} \cdot(3 \hat{\mathbf{i}}+2 \hat{\mathbf{k}})=\alpha$, then $\alpha=\ldots$

The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes $x+2 y+3 z=4$ and $4 x+3 y+2 z=1$ are $\ldots \ldots$

If line $\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other then $\lambda=$ ............

Which of the following can not be the direction cosines of a line?

If lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}$ intersect each other, then $\lambda=\ldots \ldots$

Equations of planes parallel to the plane $x-2 y+2 z+4=0$ which are at a distance of one unit from the point $(1,2,3)$ are ............

If $P(6,10,10), Q(1,0,-5), R(6,-10, \lambda)$ are vertices of a triangle right angled at $Q$, then value of $\lambda$ is ............

If the foot of the perpendicular drawn from the point $(0,0,0)$ to the plane is $(4,-2,-5)$ then the equation of the plane is .............

If $G(3,-5, r)$ is centroid of triangle $A B C$ where $A(7,-8,1), B(p, q, 5)$ and $C(q+1,5 p, 0)$ are vertices of a triangle then values of $p, q, r$ are respectively ......

The angle between lines $\frac{x-2}{2}=\frac{y-3}{-2}=\frac{z-5}{1}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-5}{2}$ is ............

If the line passes through the points $P(6,-1,2), Q(8,-7,2 \lambda)$ and $R(5,2,4)$ then value of $\lambda$ is ...........

The co-ordinates of the foot of perpendicular drawn from origin to the plane $2 x-y+5 z-3=0$ are $\ldots \ldots$

The equation of the plane passing through the point $(-1,2,1)$ and perpendicular to the line joining the points $(-3,1,2)$ and $(2,3,4)$ is