Let 0 < z < y < x be three real numbers such that (1)/(x), (1)/(y),...

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

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

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

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

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

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

$⇒ x + z = 4\sqrt{2}$

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

$= 150$

Therefore, this is the required answer.