Explanation

(a)
It is obvious that the given line and plane are parallel
Given point on the line is $A(2,\ - 2,\ 3)$
$B(0,\ 0,\ 5)$ is a point on the plane
$\therefore\ \overrightarrow{AB} = (2 - 0)\widehat{i} + ( - 2 - 0)\widehat{j} + (3 - 5)\widehat{k}$
Then distance of $B$ from the plane = projection of $\overrightarrow{AB}$ on vector $\widehat{i} + 5\widehat{j} + \widehat{k}$
$p = \left| \frac{\left( 2\widehat{i} - 2\widehat{j} - 2\widehat{k} \right) \bullet (\widehat{i} + 5\widehat{j} + \widehat{k})}{\sqrt{1 + 25 + 1}} \right|$
$= \left| \frac{2 - 10 - 2}{\sqrt{27}} \right| = \frac{10}{3\sqrt{3}}$