1
JEE Advanced 2024 Paper 1 Online
+3
-1

Let $\gamma \in \mathbb{R}$ be such that the lines $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O=(0,0,0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$.

Match each entry in List-I to the correct entry in List-II.

List-I List-II
(P) $\gamma$ equals (1) $-\hat{i} - \hat{j} + \hat{k}$
(Q) A possible choice for $\hat{n}$ is (2) $\sqrt{\frac{3}{2}}$
(R) $\overrightarrow{OR_1}$ equals (3) $1$
(S) A possible value of $\overrightarrow{OR_1} \cdot \hat{n}$ is (4) $\frac{1}{\sqrt{6}} \hat{i} - \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}$
(5) $\sqrt{\frac{2}{3}}$

The correct option is :
A
$(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
B
$(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
C
$(\mathrm{P}) \rightarrow(3) \quad$ (Q) $\rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad$ (S) $\rightarrow(5)$
D
$(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(1) \quad(\mathrm{R}) \rightarrow(4) \quad$ (S) $\rightarrow(5)$
2
JEE Advanced 2023 Paper 1 Online
+3
-1
Let $\ell_1$ and $\ell_2$ be the lines $\vec{r}_1=\lambda(\hat{i}+\hat{j}+\hat{k})$ and $\vec{r}_2=(\hat{j}-\hat{k})+\mu(\hat{i}+\hat{k})$, respectively. Let $X$ be the set of all the planes $H$ that contain the line $\ell_1$. For a plane $H$, let $d(H)$ denote the smallest possible distance between the points of $\ell_2$ and $H$. Let $H_0$ be a plane in $X$ for which $d\left(H_0\right)$ is the maximum value of $d(H)$ as $H$ varies over all planes in $X$.

Match each entry in List-I to the correct entries in List-II.

List - I List - II
(P) The value of $d\left(H_0\right)$ is (1) $\sqrt{3}$
(Q) The distance of the point $(0,1,2)$ from $H_0$ is (2) $\frac{1}{\sqrt{3}}$
(R) The distance of origin from $H_0$ is (3) 0
(S) The distance of origin from the point of intersection of planes $y=z, x=1$ and $H_0$ is (4) $\sqrt{2}$
(5) $\frac{1}{\sqrt{2}}$

The correct option is:
A
$$(P) \rightarrow(2) \quad(Q) \rightarrow(4) \quad(R) \rightarrow(5) \quad(S) \rightarrow(1)$$
B
$$(P) \rightarrow(5) \quad(Q) \rightarrow(4) \quad(R) \rightarrow(3) \quad(S) \rightarrow(1)$$
C
$$(P) \rightarrow(2) \quad(Q) \rightarrow(1) \quad(R) \rightarrow(3) \quad(S) \rightarrow(2)$$
D
$$(P) \rightarrow(5) \quad(Q) \rightarrow(1) \quad(R) \rightarrow(4) \quad(S) \rightarrow(2)$$
3
JEE Advanced 2017 Paper 2 Offline
+3
-1
The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y $$-$$ 2z = 5 and 3x $$-$$ 6y $$-$$ 2z = 7 is
A
14x + 2y $$-$$ 15z = 1
B
$$-$$14x + 2y + 15z = 3
C
14x $$-$$ 2y + 15z = 27
D
14x + 2y + 15z = 31
4
JEE Advanced 2016 Paper 2 Offline
+3
-1
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 $${x \over 1} = {y \over 2} = {z \over 1}$$ is
A
$$x+y-3z=0$$
B
$$3x+z=0$$
C
$$x-4y+7z=0$$
D
$$2x-y=0$$
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