14. (1999 National High School Competition Question) 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 $\qquad$
14. Fill $\frac{9}{4} \sqrt{3}$. Reason: From the problem, $A H \perp$ plane $S B C$. Draw $B H \perp S C$ intersecting $S C$ 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 $S$ on plane $A B C$ be $O$, then $S O \perp$ plane $A B C$. By the inverse of the theorem of three perpendiculars, we know $C O \perp A B$ at $F$, similarly $B O \perp A C$, so $O$ is the orthocenter of $\triangle A B C$. Since $\triangle A B C$ is an equilateral triangle, $O$ is 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, $E F \perp A B$, so $\angle E F C$ is the plane angle of the dihedral angle $H-A B-C$, hence $\angle E F C=30^{\circ}, O C=S C \cdot \cos 60^{\circ}=2 \sqrt{3} \cdot \frac{1}{2}=$ $\sqrt{3}, S O=\sqrt{3} \cdot \tan 60^{\circ}=\sqrt{3} \cdot \sqrt{3}=3$. Also, $O C=\frac{\sqrt{3}}{3} A B$, so $A B=\sqrt{3} O C=\sqrt{3} \cdot \sqrt{3}=3$, thus $V_{S-A B C}=\frac{1}{3} \cdot$ $\frac{\sqrt{3}}{4} \cdot 3^{2} \cdot 3=\frac{9}{4} \sqrt{3}$