If the distance between the plane $\mathrm { Ax } - 2 \mathrm { y } + \mathrm { z } = \mathrm { d }$ 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 $\sqrt { 6 }$, then $| d |$ is