Answer:
Option A
Explanation:
L1:x−12=y−0−1=z−(−3)1
Normal of plane p:n= [ˆiˆjˆk71235−6]
= ˆi(−16)−ˆj(−42−6)+ˆk(32)
=−16ˆi+48ˆj+32ˆk
DR's of normal n= ˆi−3ˆj−2ˆk
Point of intersection of L1 and L2
⇒ 2K1+1=K2+4and−k1=k2−3
∴ k1= 2 and k2 =1
\therefore point of intersection (5,-2,-1)
\therefore Equation of plane ,
1.(x-5)-3 (y+2)-2(z+1)=0
\Rightarrow x-3y-2z-13=0
\Rightarrow x-3y-2z=13
\therefore a=1, b=-3, c=-2, d=13