The equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $x + 2y - 5z + 9 = 0$ 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 two planes be given by $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Based on the given information,find the equation of the plane passing through the intersection of $P_1$ and $P_2$ and the point $(3, 2, 1)$.

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

If the distance between the plane $Ax-2y+z=d$ 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 equation of the plane passing through the points $(3, 2, 2)$ and $(1, 0, -1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{3}$ is: