Challenging Volume Calculation

The slant height of the pyramid is $\sqrt{5^2 + 12^2} = 13$ , so $\angle ADB = \angle BDC = \tan^{-1} \frac{13}{5}$ , which means $\cos \angle ADB = \cos \angle BDC = \frac{5}{\sqrt{194}}$ . $\angle ADC$ is a right angle, so $\cos \angle ADC = 0$ .

Let the coordinates of $D$ be $(p, q, r)$ . Then $\vec{DB} = (-p, -q, -r)$ , $\vec{DA} = (-p - 1, 7 - q, -r)$ , and $\vec{DC} = (-p + 6, -q, -r)$ .

Using $\cos \theta = \frac{u \bullet v}{|u||v|}$ , we have:

$\text{ } \cos \angle ADB = \frac{5}{\sqrt{194}} = \frac{p(p + 1) + q(q - 7) + r^2}{\sqrt{p^2 + q^2 + r^2}\sqrt{(p + 1)^2 + (q - 7)^2 + r^2}}$

$\text{ } \cos \angle BDC = \frac{5}{\sqrt{194}} = \frac{p(p - 6) + q^2 + r^2}{\sqrt{p^2 + q^2 + r^2}\sqrt{(p - 6)^2 + q^2 + r^2}}$

$\text{ } \cos \angle ADC = 0 = (p - 6)(p + 1) + q(q - 7) + r^2$

which solves numerically to $(p, q, r) \approx (0.369, 0.185, -2.994)$ , which means the height of the submerged tetrahedron is $h \approx 2.994$ .

Since $\vec{BA} = (-1, 7, 0)$ and $\vec{BC} = (6, 0, 0)$ , $\triangle ABC$ has an area of $A = \frac{1}{2}\begin{Vmatrix} -1 & 7 \\ 6 & 0 \end{Vmatrix} = 21$ .

Therefore, the volume of the submerged tetrahedron is $V \approx \frac{1}{3} \cdot 21 \cdot 2.994 \approx \boxed{20.959}$ .