jmc

Let $△ABC$ (or the triangle with sides $12\sqrt{3}$, $13\sqrt{3}$, $13\sqrt{3}$) be the base of our tetrahedron. We set points $C$ and $D$ as $(6\sqrt{3},0,0)$ and $(-6\sqrt{3},0,0)$, respectively. Using Pythagoras, we find $A$ as $(0,\sqrt{399},0)$. We know that the vertex of the tetrahedron ($P$) has to be of the form $(x,y,z)$, where $z$ is the altitude of the tetrahedron. Since the distance from $P$ to points $A$, $B$, and $C$ is $\frac{\sqrt{939}}{2}$, we can write three equations using the distance formula: $$\begin{array}{rl}{x}^2+(y-\sqrt{399}{)}^2+z^2& =\frac{939}{4}\\ (x-6\sqrt{3}{)}^2+y^2+z^2& =\frac{939}{4}\\ (x+6\sqrt{3}{)}^2+y^2+z^2& =\frac{939}{4}\end{array}$$ Subtracting the last two equations, we get $x=0$. Solving for $y,z$ with a bit of effort, we eventually get $x=0$, $y=\frac{291}{2\sqrt{399}}$, $z=\frac{99}{\sqrt{133}}$. Since the area of a triangle is $\frac{1}{2}\cdot bh$, we have the base area as $18\sqrt{133}$. Thus, the volume is $V=\frac{1}{3}\cdot 18\sqrt{133}\cdot \frac{99}{\sqrt{133}}=6\cdot 99=\boxed{594}$.