5. Given a sphere with a radius of 6. Then the maximum volume of a regular tetrahedron inscribed in the sphere is ( ).
(A) $32 \sqrt{3}$
(B) $54 \sqrt{3}$
(C) $64 \sqrt{3}$
(D) $72 \sqrt{3}$
5. C.

Let the angle between the height and a lateral edge of a regular tetrahedron be $\theta$. Then the length of the lateral edge $l=12 \cos \theta$, and the height $h=12 \cos ^{2} \theta$.
Thus, the side length of the equilateral triangle at the base is
$$
a=12 \sqrt{3} \sin \theta \cdot \cos \theta \text {. }
$$

Therefore, the volume sought is
$$
\begin{array}{l}
V=432 \sqrt{3} \cos ^{4} \theta \cdot \sin ^{2} \theta \\
=216 \sqrt{3} \cos ^{2} \theta \cdot \cos ^{2} \theta \cdot 2 \sin ^{2} \theta \\
\leqslant 216 \sqrt{3}\left(\frac{\cos ^{2} \theta+\cos ^{2} \theta+2 \sin ^{2} \theta}{3}\right)^{3} \\
=64 \sqrt{3} .
\end{array}
$$

The equality holds when $\tan ^{2} \theta=\frac{1}{2}$.