Question

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 $1$ with List $2$ and select the correct answer using the code given below the lists:

Answer 1

The plane perpendicular to $P_1$ and $P_2$ has direction ratios of the normal.
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 7& 1& 2\\ 3& 5& -6\end{array}\right|=-16\hat{i}+48\hat{j}+32\hat{k}$....... $\left(1\right)$
For point of intersection of lines
$(2{\lambda }_1+1,-{\lambda }_1,{\lambda }_1-3)\equiv ({\lambda }_2+4,{\lambda }_2-3,2{\lambda }_2-3)$
$\Rightarrow 2{\lambda }_1+1={\lambda }_2+4$ or $2{\lambda }_1-{\lambda }_2=3$
$-{\lambda }_1={\lambda }_2-3$ or ${\lambda }_1+{\lambda }_2=3$
$\Rightarrow {\lambda }_1=2,{\lambda }_2=1$
$\therefore$ Point is $(5,-2,-1)\cdots$... $\left(2\right)$
From (1) and (2), the required plane is
$-1(x-5)+3(y+2)+2(z+1)=0$
$or-x+3y+2z=-13$
$x-3y-2z=13$
$\Rightarrow a=1,b=-3,c=-2,d=13$