Let the line passing through the points $P\left(2,-1,2\right)$ and $Q\left(5,3,4\right)$ meet the plane $x-y+z=4$ at the point $R$. Then the distance of the point $R$ from the plane $x+2y+3z+2=0$ measured parallel to the line $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ is

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,

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda$

Now let point $R$ be $\left(3\lambda +2,4\lambda -1,2\lambda +2\right)$

Given, $R$ lies on plane $x-y+z=4$

$\therefore 3\lambda +2-4\lambda +1+2\lambda +2=4$

$\Rightarrow \lambda =-1$

$\therefore R\left(-1,-5,0\right)$

Now let line $SR$ be :$\frac{x+1}{2}=\frac{y+5}{2}=\frac{z}{1}=k$ as it is parallel to $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ and passing through $\left(-1,-5,0\right)$

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

Now $S$ lies on plane : $x+2y+3z+2=0$

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

$\Rightarrow 9k-9=0\Rightarrow k=1$

Hence, $S\left(1,-3,1\right)$

So, distance $SR=\sqrt{4+4+1}=3$