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 :