6. 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 dihedral angle $H-AB-C$ is $30^{\circ}$, and $SA=2\sqrt{3}$. Then the volume of the tetrahedron $S-ABC$ is
$$
\frac{9}{4} \sqrt{3}
$$
6. 【Analysis and Solution】As shown in the figure, given that $A H \perp$ plane $S B C$, draw $B H \perp S C$ at $E$. By the theorem of three perpendiculars, we know that $S C \perp A E, S C \perp A B$. Therefore, $S C \perp$ 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 that $C O \perp A B$ at $F$. Similarly, $B O \perp A C$. Hence, $O$ is the orthocenter of $\triangle A B C$.

Since $\triangle A B C$ is an equilateral triangle, $O$ is the center of $\triangle A B C$, thus $S A=$ $S B=S C=2 \sqrt{3}$.

Since $C F \perp A B, C F$ is the projection of $E F$ on plane $A B C$, and by the theorem of three perpendiculars, $E F \perp A B$.
Therefore, $\angle E F C$ is the plane angle of the dihedral angle $H-A B-C$.
Hence $\angle E F C=30^{\circ} \Rightarrow O C=S C \cos 60^{\circ}=2 \sqrt{3} \times \frac{1}{2}=\sqrt{3}$.
$$
\begin{array}{l}
S O=\sqrt{3} \tan 60^{\circ}=\sqrt{3} \times \sqrt{3}=3 . \\
\text { Also } O C=\frac{\sqrt{3}}{3} A B, \therefore A B=\sqrt{3} O C=\sqrt{3} \times \sqrt{3}=3 . \\
\therefore V_{S-A B C}=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 3^{2} \times 3=\frac{9}{4} \sqrt{3} .
\end{array}
$$