Tetrahedral Slice

Let $p=||AP||,q=||AQ||,r=||AR||$ . Using basic trigonometry, we find that the area of $\Delta APR$ is $\sin(\frac{\pi}{3})\frac{pr}{2}=\frac{\sqrt3}{4}pr$ . Writing analogous equations for the other two triangles and multiplying them together, we find that $\left(\frac{\sqrt 3}{4}\right)^3(pqr)^2=10\sqrt{3}(14^2-61)$ so $pqr=120\sqrt{2}$ . Finally, scaling the volume formula $V=\frac{a^3}{6\sqrt{2}}$ for a regular tetrahedron, we can conclude that $V=\frac{pqr}{6\sqrt{2}}=\boxed{20}$ .

A charming and very well-crafted problem! Thank you for sharing!