Let a plane $P$ pass through the point $A(3, 7, -7)$ and contain the line $L: \frac{x - 2}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$. If the distance of the plane $P$ from the origin is $d$, then $d^2$ is

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$,and $\pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$,then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$

Find the distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$.

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

Consider the lines $L_1: \frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}$,$L_2: \frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}$ and the planes $P_1: 7x+y+2z=3$,$P_2: 3x+5y-6z=4$. Let $ax+by+cz=d$ be the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$,and perpendicular to planes $P_1$ and $P_2$. Match List-$I$ with List-$II$ and select the correct answer using the code given below the lists:

List-$I$	List-$II$
$P. \quad a =$	$1. \quad 13$
$Q. \quad b =$	$2. \quad -3$
$R. \quad c =$	$3. \quad 1$
$S. \quad d =$	$4. \quad -2$

Codes: $P \quad Q \quad R \quad S$