If the image of the point P(1, -2, 3) in the plane 2x + 3y - 4z + 22 = 0 measured parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is Q, then PQ is equal to

$2\sqrt{42}$

Let R be the point where the line passing through point P(1, -2, 3) and parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ intersects the plane 2x + 3y - 4z + 22 = 0. Then, PQ = 2PR.

The equations of the line passing through P(1, -2, 3) and parallel tp $\frac{x}{1}=\frac{y}{4}=\frac{z}{3}$ is $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{3}$. Suppose it intersects the plane at R. The coordinates of R are given by 

$\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{3}=λ $ or, $ (λ+1, 4λ -2, 5λ +3)$

It lies on the plane 2x + 3y - 4z + 22= 0.

$∴ 2λ + 2 + 12 λ - 6 - 20 λ - 12 + 22= 0 ⇒ -6 + 6 = 0 ⇒λ = 1 $

So, the coordinates of R are (2 , 2, 8).

$∴ PR = \sqrt{(2-1)^2 +(2+2)^2 +(8-3)^2}= \sqrt{42}$

Hence, $PQ = 2PR = 2\sqrt{42}$.