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 :