If the three planes x = 5, 2x - 5ay + 3z - 2 = 0 and 3bx + y - 3z = 0 contain a common line, then (a, b) is equal to :

$\left(\frac{1}{5}, -\frac{8}{15}\right)$

The line of intersection of first two planes is 

$\frac{x-5}{0}= \frac{y}{-3}= \frac{z+8/3}{-5a}$

It must lie on third plane.

$∴ 3b × 0 + (-3) × 1 + (-3) (-5a) = 0 $

and, $ 3b × 5 + 0 × 1 + (-3) (-8/3) = 0 $

$⇒ a = \frac{1}{5}$ and $ 15b + 8 = 0 ⇒ a = \frac{1}{5}$ and $ b = -\frac{8}{15}$