Example 1 Given that the base of the tetrahedron $S-ABC$ is an equilateral triangle, the projection of point $A$ on the side face $SBC$ is $H$, which is the orthocenter of $\triangle SBC$, the plane angle of the dihedral angle $H-AB-C$ is $30^{\circ}$, and $SA=2\sqrt{3}$. Find the volume of the tetrahedron $S-ABC$.
(1999 National High School League Question)
As shown in Figure 22-2, from the given conditions, we know that $A H \perp$ plane $S B C$. Draw $B H \perp S C$ at $E$, then by the theorem of three perpendiculars, we know that $S C \perp$ plane $A B E$.

Let the projection of $S$ on plane $A B C$ be $O$, then $S O \perp$ plane $A B C$. By the inverse theorem of the three perpendiculars, we know that $C O \perp A B$ at $F$. Similarly, $B O \perp A C$. Therefore, $O$ is the center of $\triangle A B C$, hence $S A=S B=S C=2 \sqrt{3}$.

Also, $C F \perp A B, C F$ is the projection of $E F$ on plane $A B C$, by the theorem of three perpendiculars, we know that $E F \perp A B$, so $\angle E F C$ is the plane angle of the dihedral angle $H-A B-C$, thus $\angle E F C=30^{\circ}, O C=S C \cdot \cos 60^{\circ}=\sqrt{3}, S O=O C \cdot \tan 60^{\circ}=3$. Also, $O C=$ $\frac{\sqrt{3}}{3} A B$, then $A B=\sqrt{3} O C=3$, so, $V_{S-A B C}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot 3^{2} \cdot 3=\frac{9}{4} \sqrt{3}$.