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 $xy+yz+zx=\frac{3}{\sqrt{2}}xyz$, then $3(x+y+z{)}^2$ is equal to

Given that $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in AP and $x,\sqrt{2}y,z$ are in GP.

As given, $\frac{2}{y}=\frac{1}{x}+\frac{1}{y} .......\left(\mathrm{i}\right)$
Also, $2y^2=xz .......\left(\mathrm{ii}\right)$

Also given that $xy+yz+zx=\frac{3}{\sqrt{2}}xyz$
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{\sqrt{2}} .....\left(\mathrm{iii}\right)$
From $\left(i\right)$ and $\left(iii\right)$ we get $\frac{3}{y}=\frac{3}{\sqrt{2}}$

$y=\sqrt{2} .....\left(\mathrm{iv}\right)$

Now from $\left(ii\right)$ $xz=4 .......\left(\mathrm{v}\right)$

Now using $\left(ii\right), \left(iv\right) \text{and} \left(v\right)$

$\Rightarrow x+z=4\sqrt{2}$

Hence $3(x+y+z{)}^2=3(\sqrt{2}+4\sqrt{2}{)}^2$

$=150$

Therefore, this is the required answer.