Example 3 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 $SA = 2\sqrt{3}$. Then, the volume of the tetrahedron $S-ABC$ is
Analysis: The given segment length seems insufficient. From the basic conclusion 7, we know that the three pairs of opposite edges are mutually perpendicular, so the projection $O$ of $S$ on plane $A B C$ is also the orthocenter of $\triangle A B C$. Furthermore, since $\triangle A B C$ is an equilateral triangle, $O$ is the circumcenter of $\triangle A B C$. Therefore, from the basic conclusion 5, we know that $S A=S B=S C$. Connect $B H$ and extend it to intersect $S C$ at $E$, then connect $A E$, so $S C \perp$ plane $E A B$. Draw $E F \perp A B$, connect $C F$, and we know that $\angle E F C$ is the plane angle of the dihedral angle $H-A B-C$, i.e., $\angle E F C=30^{\circ}$. Therefore, $\angle S C O=60^{\circ}$. According to this, solve for $O S$ and $O C$, and it is easy to calculate the required volume as $\frac{9 \sqrt{3}}{4}$.