Let $A_1$, $A_2$, $A_3$, … , $A_n$ be a set of points, and let $G=\displaystyle\frac{A_1+A_2+\ldots+A_n}{n}$.
Now take any circle centered at $G$. (If you're working in 3 dimensions, take a sphere centered at $G$ instead.) For any point $P$ on this circle, the value $A_1P^2+A_2P^2+\ldots+A_nP^2$ is constant.
To prove this, first note that, for any point $X$, we have $\vec{A_1X}+\vec{A_2X}+\ldots+\vec{A_nX}=n\vec{XG}$.
From this, you can conclude that $\vec{A_1G}+\vec{A_2G}+\ldots+\vec{A_nG}=0$.
So, for any point $X$, we have the following:
$$A_1X^2+A_2X^2+\ldots+A_nX^2=\left|\vec{A_1X}\right|^2+\left|\vec{A_2X}\right|^2+\ldots+\left|\vec{A_nX}\right|^2$$
$$=\left|\vec{A_1G}+\vec{XG}\right|^2+\left|\vec{A_2G}+\vec{XG}\right|^2+\ldots+\left|\vec{A_nG}+\vec{XG}\right|^2$$
$$=n\cdot\left|\vec{XG}\right|^2+2\cdot\vec{XG}\cdot\left(\vec{A_1G}+\vec{A_2G}+\ldots+\vec{A_nG}\right)+\left|\vec{A_1G}\right|^2+\left|\vec{A_2G}\right|^2+\ldots+\left|\vec{A_nG}\right|^2$$
$$=n\left|\vec{XG}\right|^2+\left|\vec{A_1G}\right|^2+\left|\vec{A_2G}\right|^2+\ldots+\left|\vec{A_nG}\right|^2$$
$$=n\cdot {XG}^2+A_1G^2+A_2G^2+\ldots+A_nG^2$$
So, when $XG$ is constant, $A_1X^2+A_2X^2+\ldots+A_nX^2$ is constant.
Sources: http://www.cut-the-knot.org/triangle/medians.shtml
Side note if you like physics:
Imagine $n$ Hookean springs, all with the same spring constant. The first spring connects $A_1$ to $X$, the second spring connects $A_2$ to $X$, and etc. In this setup, points $A_1$ through $A_n$ are fixed, but point $X$ is free to move. The potential energy of this setup is proportional to $A_1X^2+A_2X^2+\ldots+A_nX^2$. Since nature likes to minimize potential energy, this setup will come to rest when $A_1X^2+A_2X^2+\ldots+A_nX^2$ is at it's minimum. This setup will also come to rest when all the forces sum to $0$. At any point $X$, the sum of the forces will be $k\cdot\left(\vec{A_1}+\vec{A_2}+\ldots+\vec{A_n}\right)=k\cdot n\cdot XG$, where $k$ is the spring constant of each spring. This vector is sort of the opposite of a gradient vector, in the sense that it points to the direction of greatest decrease. If point $X$ moves perpendicular to this vector, there will be no change in the value of $A_1X^2+A_2X^2+\ldots+A_nX^2$. Ergo, you can move $X$ along a circle centered at $G$, and the value $A_1X^2+A_2X^2+\ldots+A_nX^2$ will remain constant.
Let's use an orthogonal reference, with $O$ at the origin, $\mathbf a$ along the first axis (unit vector $\hat i$), $\mathbf b$ in the $\hat i\hat j$ plane, and $\mathbf c$ with components along all three axes. Since we know the angle between vectors is $\pi/3$, we can write
$$\mathbf a=x\hat i\\
\mathbf b=x\cos\frac{\pi}{3}\hat i+x\sin\frac{\pi}{3}\hat j\\
\mathbf c=\alpha \hat i+ \beta \hat j+\gamma\hat k
$$
The distance from $C$ to the plane $ABO$ is $\gamma$. We can now use $$\mathbf a\cdot\mathbf c=x^2\cos\frac{\pi}{3}\\ \mathbf b\cdot\mathbf c=x^2\cos\frac{\pi}{3}\\ \mathbf c\cdot\mathbf c=x^2$$ to calculate all the components.
From the first equation $x\alpha=x^2\cos\frac{\pi}{3}$ or $$\alpha=x/2$$ From the second equation $$x\frac{1}{2}\alpha+x\frac{\sqrt 3}{2}\beta=x^2\frac{1}{2}$$ This yields $$\beta=\frac{x}{2\sqrt 3}$$
The third equation is $$\alpha^2+\beta^2+\gamma^2=x^2$$
Plugging in the previous values for $\alpha$ and $\beta$ we get $$\gamma=x\sqrt\frac{2}{3}$$
Best Answer
I happened to have a suitable image on my laptop (for my freshman course), so I couldn't resist
It is supposed to be a methane molecule as opposed to a tetrahedron, so think of the hydrogen atoms (green) as vertices of the tetrahedron. As you see, this is the second arrangement in Old John's (+1) answer. One vertex is at the point $(1,1,1)$ and the other three at the points $(\pm1,\pm1,\pm1)$ with exactly two minus signs occuring. In other words, four selected corners of a cube.
I second Old John's opinion that this is very well suited for 3D-rotations, because you surely know the effect of a rotation on the cube...