Let the plane P contain the line 2x + y - z - 3 = 0 = 5x - 3y + 4z...

Let the Plane containing the line $2x + y - 3 - 3 = 0 = 5x - 3y + 4z + 9$ is given by,

$⇒ 2x + y - z - 3 + \lambda\left(5x - 3y + 4z + 9\right) = 0$

$⇒ x\left(2 + 5\lambda\right) + y\left(1 - 3\lambda\right) + z\left(4\lambda - 1\right) + 9\lambda - 3 = 0$

Also, this plane is parallel to the line

$\dfrac{x + 2}{2} = \dfrac{3 - y}{- 4} = \dfrac{z - 7}{5}$

$⇒ \dfrac{x + 2}{2} = \dfrac{y - 3}{4} = \dfrac{z - 7}{5}$

Now normal of the plane will be perpendicular to the line, so using perpendicular condition we get,

$∴ \left(2 + 5\lambda\right)\left(2\right) + \left(1 - 3\lambda\right)\left(4\right) + \left(4\lambda - 1\right)5 = 0$

$⇒ 4 + 10\lambda + 4 - 12\lambda + 20\lambda - 5 = 0$

$⇒ 18\lambda = - 3 ⇒ \lambda = \dfrac{- 1}{6}$

So, equation of the Plane will be,

$P: \dfrac{7}{6}x + \dfrac{3}{2}y - \dfrac{5}{3}z - \dfrac{9}{2} = 0$

$⇒ 7x + 9y - 10z - 27 = 0$

Now to find $AB$,

Let $\dfrac{x - 8}{- 3} = \dfrac{y + 1}{4} = \dfrac{z + 19}{12} = k$

So, the point $B = \left(– 3k + 8, 4k - 1, 12k - 19\right)$

And $B$ lies on plane $P$.

$7\left(– 3k + 8\right) + 9\left(4k - 1\right) - 10\left(12k - 19\right) = 27$

$⇒ – 21k + 56 + 36k - 9 - 120k + 190 = 27$

$⇒ - 105k = – 210$

$⇒ k = 2$

$∴ B = \left(2, 7, 5\right)$

Hence, Distance is $AB = \sqrt{36 + 64 + 576}$

$= \sqrt{676} = 26$

Hence this is the required answer.