Call Lagrange

Using Lagrange multipliers: $2x=\lambda yz, 2y=\lambda xz, 2z=\lambda xy \Rightarrow x^2=\frac{\lambda}{2}xyz, y^2=\frac{\lambda}{2}xyz, z^2=\frac{\lambda}{2}xyz \Rightarrow x^2=y^2=z^2=(\frac{\lambda}{2})^2$ . So, $y=\pm x$ and $z=\pm x$ . This leads to $x^3=\pm 1$ . So, $x=1$ when $y=z=\pm 1$ , and $x=-1$ when $y=-z=\pm 1$ . Putting this together, we get the points (1, 1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1).

The distance from the surface to the origin is given by $\sqrt{x^2+y^2+z^2}$ . By AM-GM inequality, we have:

$\begin{aligned} x^2 + y^2 + z^2 & \ge 3\sqrt[3]{(xyz)^2} = 3 \end{aligned}$

Equality occurs when $x^2=y^2=z^2=1$ . With $xyz=1$ , there are two possibilities, all $x, y, z =1$ or one of them is 1 and the other two are -1. Altogether 4 points $(1,1,1)$ , $(-1,-1,1)$ , $(-1,1-1)$ , and $(1,-1,-1)$ . Therefore, the sum of squares of all coordinates is $4\times 3=\boxed{12}$ .