Explanation

(a)
Given lines are
$\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{\mathcal{z} + 2}{1} = r_{1}$ (say)
and $\frac{x + 3}{- 36} = \frac{y - 3}{2} = \frac{\mathcal{z} - 6}{4} = r_{2}$ (say)
$\therefore x = 3r_{1} + 5 = - 36r_{2} - 3$
$y = - r_{1} + 7 = 3 + 2r_{2}$
and $z = r_{1} - 2 = 4r_{2} + 6$
On solving, we get
$x = 21,\ y = \frac{5}{3},\mathcal{z} = \frac{10}{3}$