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 plane passing through the point of intersection of the lines $L_1$ and $L_2$, and perpendicular to the planes $P_1$ and $P_2$.

Match $\text{List - I}$ with the $\text{List - II}$ and select the correct answer using the code given below the lists :
$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{P}& a=& 1.& 13\\ \text{Q}& b=& 2.& -3\\ \text{R}& c=& 3.& 1\\ \text{S}& d=& 4.& -2\end{array}$

• A. $\text{(P)}\to \text{(3)},\text{(Q)}\to \text{(2)},\text{(R)}\to \text{(4)},\text{(S)}\to \text{(1)}$
• B. $\text{(P)}\to \text{(1)},\text{(Q)}\to \text{(3)},\text{(R)}\to \text{(4)},\text{(S)}\to \text{(2)}$
• C. $\text{(P)}\to \text{(3)},\text{(Q)}\to \text{(2)},\text{(R)}\to \text{(1)},\text{(S)}\to \text{(4)}$
• D. $\text{(P)}\to \text{(2)},\text{(Q)}\to \text{(4)},\text{(R)}\to \text{(1)},\text{(S)}\to \text{(3)}$

Answer 1

The correct option is A $\text{(P)}\to \text{(3)},\text{(Q)}\to \text{(2)},\text{(R)}\to \text{(4)},\text{(S)}\to \text{(1)}$

Any point on line $L_1(2\lambda +1,-\lambda ,\lambda -3)$
Any point on line $L_2(\mu +4,\mu -3,2\mu -3)$
For point of intersection of $L_1$ and $L_2$
$\Rightarrow 2\lambda +1=\mu +4,-\lambda =\mu -3,\lambda -3=2\mu -3$
So $\lambda =2,\mu =1$
$\therefore$ Point of intersection is $(5,-2,-1)$
Let normal of palne $ax+by+cz=d$ be $\overrightarrow{n}$
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 7& 1& 2\\ 3& 5& -6\end{array}\right|$
$\Rightarrow \overrightarrow{n}=-16\hat{i}+48\hat{j}+32\hat{k}$
$\therefore \overrightarrow{n}=-\hat{i}+3\hat{j}+2\hat{k}$
Hence equation of plane is
$\Rightarrow -1(x-5)+3(y+2)+2(z+1)=0$
$\therefore x-3y-2z=13$
So $a=1,b=-3,c=-2,d=13$