Question

The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,–4,–5)$ and $R(2,–3,1)$ and the plane
$2x+y+z=7$, is equal to

Answer 1

$QR=\frac{x-2}{-1}=\frac{y+3}{1}=\frac{z-1}{6}$
Let point of intersection be $S(-\lambda +2,\lambda -3,6\lambda +1)$
Above point lies on the given plane
$2(-\lambda +2)+\lambda -3+6\lambda +1=7\Rightarrow \lambda =1$
So, $S(1,-2,7)$
and given $P(3,4,4)$
Now, $PS=\sqrt{2^2+6^2+3^2}=7$