Example 3 Given that the base of the tetrahedron $S \cdots A B C$ is an equilateral triangle, point $A$'s projection $H$ on the side face $SBC$ is the centroid of $\triangle S B C$, … the dihedral angle $H-A B-C$ is equal to $30^{\circ}, S A=2 \sqrt{3}$. Then, the volume of the tetrahedron $S$ $A B C$ is $\qquad$
(1999, National High School League)
Solution: As shown in Figure 3, given $A H \perp$ plane $S B C$. Draw $B H$ $\perp S C$ at $E$.

By the theorem of three perpendiculars, we know $S C \perp A E, S C \perp A B$, hence $S C \perp$ plane $A B E$. Let the projection of point $S$ on plane $A B C$ be $O$, then $S O \perp$ plane $A B C$. By the converse of the theorem of three perpendiculars, we know $C O \perp A B$ at $F$. Similarly, $B O \perp A C$. Therefore, point $O$ is the orthocenter of $\triangle A B C$. Since $\triangle A B C$ is an equilateral triangle, point $O$ is also the centroid of $\triangle A B C$, thus $S A=S B=S C=2 \sqrt{3}$. Because $C F \perp A B, C F$ is the projection of $E F$ on plane $A B C$, by the theorem of three perpendiculars, we know $E F \perp A B$. Therefore, $\angle E F C$ is the plane angle of the dihedral angle $H-A B-$ $C$, so $\angle E F C=30^{\circ}$.
$$
\begin{array}{l}
O C=S C \cos 60^{\circ}=2 \sqrt{3} \times \frac{1}{2}=\sqrt{3}, \\
S O=\sqrt{3} \tan 60^{\circ}=\sqrt{3} \times \sqrt{3}=3 . \\
\text { Also } O C=\frac{\sqrt{3}}{3} A B,
\end{array}
$$

Also $O C=\frac{\sqrt{3}}{3} A B$,
so $A B=\sqrt{3} O C=\sqrt{3} \times \sqrt{3}=3$.
Therefore, $V_{S-A B}=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 3^{2} \times 3=\frac{9}{4} \sqrt{3}$.