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