3d Geometry

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