(3) In the regular tetrahedron $D-ABC$, the side length of the base $\triangle ABC$ is 6, and the length of each lateral edge is 5. Point $I$ is the incenter of the lateral face $\triangle DAB$. Then the volume of the tetrahedron $I-ABC$ is $\qquad$ .
(3) $\frac{9}{8} \sqrt{39}$ Hint: Take the midpoint $M$ of $AB$, since $DA=DB$, then $DM \perp AB$,

and point $I$ lies on $DM$, thus $\frac{IM}{ID}=\frac{AM}{AD}=\frac{3}{5}, \frac{IM}{DM}=\frac{3}{8}$.
Let the projection of point $D$ on the base $\triangle ABC$ be $H$, then
$$
CH=\frac{2}{3} CM=\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \cdot 6=2 \sqrt{3},
$$

so
$$
\begin{array}{c}
DH=\sqrt{|DC|^{2}-|CH|^{2}}=\sqrt{13}, \\
S_{\triangle ABC}=\frac{\sqrt{3}}{4} \cdot 6^{2}=9 \sqrt{3}, \\
V_{D-ABC}=\frac{1}{3} \cdot DH \cdot S_{\triangle ABC}=3 \sqrt{39},
\end{array}
$$

Therefore, $V_{I-ABC}=\frac{3}{8} \cdot V_{D-ABC}=\frac{9}{8} \sqrt{39}$.