jmc

We want to minimize $x^2+y^2+z^2.$ We know that $xy{z}^2=2.$ Note that flipping the sign of $z$ does not change $x^2+y^2+z^2$ or $xy{z}^2,$ so we may assume that $z$ is positive. Also, from the condition $xy{z}^2=2,$ both $x$ and $y$ are positive, or both are negative. If they are both negative, we can flip the sign of both $x$ and $y.$ Thus, we may assume that $x,$ $y,$ and $z$ are all positive.

Then by AM-GM, $$\begin{array}{rl}{x}^2+y^2+z^2& =x^2+y^2+\frac{z^2}{2}+\frac{z^2}{2}\\ & \ge 4\sqrt[4]{x^2\cdot y^2\cdot \frac{z^2}{2}\cdot \frac{z^2}{2}}\\ & =4\sqrt[4]{\frac{x^2{y}^2{z}^4}{4}}\\ & =4\sqrt{\frac{xy{z}^2}{2}}\\ & =4.\end{array}$$Hence, $\sqrt{x^2+y^2+z^2}\ge 2.$

Equality occurs when $x=y=\frac{z}{\sqrt{2}}.$ Along with the condition $xy{z}^2=2,$ we can solve to get $x=1,$ $y=1,$ and $z=\sqrt{2}.$ Thus, the minimum distance is $\boxed{2}.$