Let $0 < z < y < x$ be three real numbers such that $\frac { 1 } { x } , \frac { 1 } { y } , \frac { 1 } { z }$ are in an arithmetic progression and $x , \sqrt{2} y , z$ are in a geometric progression. If $x y + y z + z x = \frac { 3 } { \sqrt { 2 } } x y z$, then $3 ( x + y + z ) ^ { 2 }$ is equal to