Let the line passing through the points P(2, - 1, 2) and Q(5, 3, 4)...

Given,

The line passing through the points $P\left(2, - 1, 2\right)$ and $Q\left(5, 3, 4\right)$,

So, equation of line $PQ$ will be,$</p><p>\dfrac{x - 2}{3} = \dfrac{y + 1}{4} = \dfrac{z - 2}{2} =$ \lambda$</p><p>Now let point$R$be \left(3\lambda + 2, 4\lambda - 1, 2\lambda + 2\right)</p><p>Given,$R$lies on plane$x - y + z = 4$</p><p>$∴ 3\lambda + 2 - 4\lambda + 1 + 2\lambda + 2 = 4$</p><p>$⇒ \lambda = - 1$</p><p>$∴ R\left(- 1, - 5, 0\right)$</p><p>Now let line$SR$be :$\dfrac{x + 1}{2} = \dfrac{y + 5}{2} = \dfrac{z}{1} =kas it is parallel$to$\dfrac{x - 7}{2} = \dfrac{y + 3}{2} = \dfrac{z - 2}{1} and passing through \left(- 1, - 5, 0\right)

$</p><p>So, the point$S: \left(2k - 1, 2k - 5, k\right)

Now $S$ lies on plane : $x + 2y + 3z + 2 = 0$

$⇒ \left(2k - 1\right)$ + \left(4k - 10\right)$+ 3k + 2 = 0$

$⇒ 9k - 9 = 0 ⇒ k = 1$

Hence, $S\left(1, - 3, 1\right)</p><p>So, distance$SR = \sqrt{4 + 4 + 1} = 3$</p>$