Three. (This question is worth 20 points) If the distance from a vertex of the base of a regular tetrahedron to the centroid of the opposite face is 4, find the maximum volume of this regular tetrahedron.


Three, as shown in the figure, for the regular triangular pyramid $P-ABC$, $G$ is the centroid of the side face $\triangle PBC$, and $AG=4$.

Extend $PG$ to intersect $BC$ at $D$. Clearly, $D$ is the midpoint of $BC$.

Draw $PO \perp$ the base $ABC$ at $O$. Then $O$ lies on $AD$, and $O$ is the centroid of $\triangle ABC$, with $OD=\frac{1}{3} AD$.

Let $\angle PDO=\alpha\left(0<\alpha<\frac{\pi}{2}\right)$, $AB=a$, and $PD=x$. Then $OD=\frac{\sqrt{3}}{6} a$, $OD=x \cos \alpha$, and $a=2 \sqrt{3} x \cos \alpha$.
Also, $PO=PD \sin \alpha=x \sin \alpha$.
$$
\begin{aligned}
V & =\frac{1}{3} S_{\triangle ABC} \cdot PO \\
& =\frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^{2} \cdot x \sin \alpha=\sqrt{3} x^{3} \cos ^{2} \alpha \sin \alpha .
\end{aligned}
$$

In $\triangle AGD$, by the cosine rule,
$$
AG^{2}=16=AD^{2}+DG^{2}-2 AD \cdot DG \cos \alpha,
$$

which is
$$
\begin{aligned}
16= & (3 x \cos \alpha)^{2}+\left(\frac{x}{3}\right)^{2} \\
& -2 \cdot 3 \cdot x \cos \alpha \cdot \frac{x}{3} \cdot \cos \alpha, \\
x^{2} & =\frac{144}{63 \cos ^{2} \alpha+1} . \\
V^{2} & =3\left(\frac{144}{63 \cos ^{2} \alpha+1}\right)^{3} \cos ^{4} \alpha \sin ^{2} \alpha \\
& =\frac{3 \times 144^{3}}{32^{2}}\left(\frac{1}{63 \cos ^{2} \alpha+1}\right)^{3} \\
& \cdot 32 \cos ^{2} \alpha \cdot 32 \cos ^{2} \alpha \cdot \sin ^{2} \alpha \\
& \leqslant \frac{4 \cdot 3^{7}}{\left(63 \cos ^{2} \alpha+1\right)^{3}}\left(\frac{32 \cos ^{2} \alpha+31 \cos ^{2} \alpha+1}{3}\right)^{3} \\
& =4 \times 3^{4},
\end{aligned}
$$
$$
V \leqslant 18 \text {. }
$$

The volume of the triangular pyramid reaches its maximum value of 18 when and only when $32 \cos ^{2} \alpha=\sin ^{2} \alpha$, i.e., $\alpha=\operatorname{arctg} 4 \sqrt{2}$.