The equation of the plane passing 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

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

Let $L$ be a line passing through the points $2 \hat{i}+3 \hat{j}+8 \hat{k}$ and $\hat{i}+6 \hat{j}+4 \hat{k}$. Let $P$ be a plane passing through $-5 \hat{i}+19 \hat{j}-14 \hat{k}$ and parallel to the vectors $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+3 \hat{k}$. If $L$ meets the plane $P$ at a point $A$,then the position vector of $A$ is:

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

Let $P_1$ and $P_2$ be two planes given by $P_1: 10x + 15y + 12z - 60 = 0$ and $P_2: -2x + 5y + 4z - 20 = 0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$

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