Three Dimensional Geometry

If the distance between the plane Ax-2y+z $=\mathrm{d}$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$ units, then $|d|$ is