Confusing square

The outer edge $BCDA$ will form an isoscelese triangle. Let put it in the xy-plane, such that its axis of symmetry coincides with the y-axis, and its base $CD$ with the x-axis. It has vertices with these coordinates:

• C will be at $\vec{v_1}=(-6\sqrt{3},0,0)$
• D will be at $\vec{v_2}=(6\sqrt{3},0,0)$

• A and B will be at $\vec{v_3}=(0,y_3,0)$
  By symmetry, the 4th vertex $\vec{v_4}$ must lie in the $yz$ -plane, so
  

• P will be at $\vec{v_4}=(0,y_4,z_4)$ .

Since $d(A,D)= 13\sqrt{3}=|v_3-v_1|=|( 6\sqrt{3},y_3,0)| =\sqrt{108+y_3^2}$ we infer that $y_3=\pm \sqrt{399}$ . Let's choose the positive one: $y_3= \sqrt{399}$

$\vec{v_4}$ is equidistant to each of the other vertices. This distance (let's name it $d$ ) is half a diagonal in the original plan, which is $d=\frac{1}{2} \sqrt{(13\sqrt{3})^2+ (12\sqrt{3})^2}=\frac{1}{2}\sqrt{939}$

Now express the squared distance to each other vertex as $d_{14}^2=d_{24}^2=108+y_4^2+z_4^2=d^2$ $d_{34}^2=(y_4-y_3)^2+z_4^2=d^2$

Setting these equal we get $108+y_4^2- (y_4-y_3)^2=0$ , which simplifies to $y_4 = y_3/2-54/ y_3$ . Using the known value for $y_3$ we get $y_4=\frac{291}{798}\sqrt{399}$

Finally, using $z_4^2=d^2- (y_4-y_3)^2$ and filling in the known values, we find $z_4=\frac{99}{\sqrt{133}}$

Numerically we have the following approximations, which makes it easier to check for mistakes:
$\vec{v_1}=(-10.39,0,0), \vec{v_2}=(10.39,0,0), \vec{v_3}=(0,19.97,0), \vec{v_4}=(0,7.284, 8.584)$
$d_{12}=20.78, d_{13}=d_{23}=22.52, d_{14}=d_{24}=15.32, d_{34}=15.32$

The volume of the piramid is given by $V=\frac{1}{3}Ah= \frac{1}{3} x_2y_3z_4$ $=\frac{1}{3} 6\sqrt{3} \sqrt{399} \frac{99}{\sqrt{133}}$ $V=\boxed{594}$