Let the equation of plane passing through the line of intersection...

Let the equation of plane passing through the intersection of two given planes:

$x + 2y + az - 2 + \lambda ( x - y + z - 3 ) = 0$

$⇒ x ( \lambda + 1 ) + y ( 2 - \lambda ) + z ( a + \lambda ) - 2 - 3\lambda = 0$

This is same as $5x - 11y + bz = 6a - 1$

$\dfrac{\lambda + 1}{5} = \dfrac{2 - \lambda}{- 11} = \dfrac{a + \lambda}{b} = \dfrac{2 + 3\lambda}{6a - 1}$

$⇒ - 11\lambda - 11 = 10 - 5\lambda$

$⇒ 6\lambda = - 21 ⇒ \lambda = \dfrac{- 7}{2}$

$\dfrac{2 - \lambda}{- 11} = \dfrac{2 + 3\lambda}{6a - 1} ⇒ \dfrac{2 + \dfrac{7}{2}}{- 11} = \dfrac{2 - \dfrac{21}{2}}{6a - 1}$$⇒ - \dfrac{1}{2} = \dfrac{- 17}{2 ( 6a - 1 )} ⇒ 6a - 1 = 17 ⇒ a = 3$

$\dfrac{2 - \lambda}{- 11} = \dfrac{a + \lambda}{b} ⇒ - \dfrac{1}{2} = \dfrac{3 - \dfrac{7}{2}}{b}$

$⇒ - \dfrac{b}{2} = - \dfrac{1}{2} ⇒ b = 1$

$∴$ point $( a, - c, c ) ≡ ( 3, - c, c )$

Distance$= \dfrac{2}{\sqrt{a}} = \dfrac{2}{\sqrt{3}}$

Plane$: 5x - 11y + z = 17$

The distance of the plane from $( 3, - c, c )$ is

\left|\dfrac{15 + 11c + c - 17}{\sqrt{147}}\right| = \dfrac{2}{\sqrt{3}}$</p><p>$⇒ | 12c - 2 | = 14$</p><p>$⇒ c = - 1, \dfrac{4}{3}$</p><p>$c = - 1\left(c \in \mathbb{Z}\right)$</p><p>$∴ \dfrac{a + b}{c} = \dfrac{3 + 1}{- 1} = - 4

Hence this is the required option.