Let 0< z <y<x be three real numbers such that 1/x, 1/y, 1/z are in an arithmetic progression and x, √ 2 y, z are in a geometric progression. If x y+y…

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