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 Column – I with Column – II and select the correct answer using the code given below the lists :

Column - I		Column - II
A)	a=	P)	13
B)	b=	Q)	-3
C)	c=	R)	1
D)	d=	S)	-2
		T)	3

Plane perpendicular to $P_1$  and $P_2$  has Direction Rations of normal
$\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 7& 1& 2\\ 3& 5& -6\end{array}\right|=-16\widehat{i}+48\widehat{j}+32\widehat{k}$    ……. (1)
For point of intersection of lines
$$\begin{gathered} (2{\lambda }_1+1,-{\lambda }_1,{\lambda }_1-3)\equiv ({\lambda }_2+4,{\lambda }_2-3,2{\lambda }_2-3) \\ \Rightarrow 2{\lambda }_1+1={\lambda }_2+4or 2{\lambda }_1-{\lambda }_2=3 \\ -{\lambda }_1={\lambda }_2-3or{\lambda }_1+{\lambda }_2=3 \\ \Rightarrow {\lambda }_1=2,{\lambda }_2=1 \end{gathered}$$
 $\therefore$ Point is (5, – 2, – 1) …………. (2)
From (1) and (2), required plane is  $-1(x-5)+3(y+2)+2(z+1)=0$  OR $-x+3y+2z=-13$

$x-3y-2z=13$
$\Rightarrow a=1,b=-3,c=-2,d=13.$