Let the equation of the plane passing through the line x - 2y - z -...

Equation of plane through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ is

\left(x - 2y - z - 5\right) + \lambda\left(x + y + 3z - 5\right) = 0$</p><p>$⇒ \left(1 + \lambda\right)x + \left(- 2 + \lambda\right) + \left(- 1 + 3\lambda\right)z + \left(- 5 - 5\lambda\right) = 0$</p><p>Also, this plane is parallel to the line$x + y + 2z - 7 = 0 = 2x + 3y + z - 2, so</p><p>\left|\begin{cases}1 + \lambda & - 2 + \lambda & - 1 + 3\lambda \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{cases}\right| = 0

$⇒ - 5\left(1 + \lambda\right) + 3\left(\lambda - 2\right) + \left(3\lambda - 1\right) = 0$

$⇒ - 5 - 5\lambda + 3\lambda - 6 + 3\lambda - 1 = 0$

$⇒ \lambda = 12$

So, equation of plane is

$13x + 10y + 35z = 65 . . . \left(1\right)$

Hence, $a = 13, b = 10, c = 35$

So, distance of the point \left(a, b, c\right)$from plane$2x + 2y - z + 16 = 0$is</p><p>$= \dfrac{26 + 20 - 35 + 16}{\sqrt{4 + 4 + 1}} = 9 units