Answer:
Option A
Explanation:
Given line,
$r= (\hat{i}-2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+2\hat{k})$ is parallel to the plane
$r. (3\hat{i}-2\hat{j}+m\hat{k})=0$
we know that, if line r= a+$\lambda$.b is parallel to plane r.n=d
then , b.n=0
$\therefore$ $(2\hat{i}+\hat{j}+2\hat{k}) .(3\hat{i}-2\hat{j}+m\hat{k}) =0 $
6-2+2m=0
m=-2