Sum of three squares

We have that $\dfrac{xz \times yz}{xy} = \dfrac{4 \times 6}{3} \Longrightarrow z^{2} = 8 \Longrightarrow z = \pm 2\sqrt{2}$ .

Then $x = \dfrac{4}{z} = \pm \sqrt{2}$ and $y = \dfrac{6}{z} = \pm \dfrac{3}{\sqrt{2}} = \pm \dfrac{3\sqrt{2}}{2}$ .

Now either all of $x,y,z$ are positive or all or negative, but in either case

$x^{2} + y^{2} + z^{2} = 2 + \dfrac{9}{2} + 8 = \dfrac{29}{2} = \boxed{14.5}$ .

$xy=3$ $(1)$

$xz=4$ $(2)$

$yz=6$ $(3)$

$(1)$ $\times$ $(2)$ $\times$ $(3)$ $= (xy)(xz)(yz) = 3(4)(6) = 72$ $\implies$ $\boxed{\color{#D61F06}x^2y^2z^2 = 72}$

Solving for $x^2$ , $y^2$ and $z^2$

$\dfrac{x^2y^2z^2}{(xy)^2} = \dfrac{72}{3^2}$ $\implies$ $\boxed{z^2 = 8}$

$\dfrac{x^2y^2z^2}{(xz)^2} = \dfrac{72}{4^2}$ $\implies$ $\boxed{y^2 = 4.5}$

$\dfrac{x^2y^2z^2}{(yz)^2} = \dfrac{72}{6^2}$ $\implies$ $\boxed{x^2 = 2}$

Adding $x^2$ , $y^2$ and $z^2$

$x^2$ $+$ $y^2$ $+$ $z^2$ $= 8 + 4.5 + 2 =$ $\boxed{\color{#3D99F6}\large14.5}$

Multiplying all three equations, we get $x^2 y^2 z^2 = 72$ . Then $\begin{aligned} x^2 &= \frac{x^2 y^2 z^2}{y^2 z^2} = \frac{72}{36} = 2, \\ y^2 &= \frac{x^2 y^2 z^2}{x^2 z^2} = \frac{72}{16} = \frac{9}{2}, \\ z^2 &= \frac{x^2 y^2 z^2}{x^2 y^2} = \frac{72}{9} = 8, \end{aligned}$ so $x^2 + y^2 + z^2 = 2 + \frac{9}{2} + 8 = \frac{29}{2}.$