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 :

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:

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

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