If the distance between the plane $$Ax-2y+z=d$$ and the plane containing the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$ and $${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}\,$$ is $$\sqrt 6 \,\,,$$ then find $$\left| d \right|.$$