3. In a regular tetrahedron $D-ABC$, it is known that the side lengths of the base $\triangle ABC$ are all 6, and the lengths of each lateral edge are all 5. $I$ is the incenter of the lateral face $\triangle DAB$. Then the volume of the tetrahedron $IABC$ is $\qquad$
3. $\frac{9 \sqrt{39}}{8}$.

Take the midpoint $M$ of $A B$.
Since $D A=D B$, then $D M \perp A B$, and point $I$ lies on side $D M$.
Thus $\frac{I M}{I D}=\frac{A M}{A D}=\frac{3}{5} \Rightarrow \frac{I M}{D M}=\frac{3}{8}$.
Let the projection of point $D$ on the base $\triangle A B C$ be $H$. Then
$$
\begin{array}{l}
C H=\frac{2}{3} C M=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 6=2 \sqrt{3} \\
\Rightarrow D H=\sqrt{|D C|^{2}-|C H|^{2}}=\sqrt{13} .
\end{array}
$$

And $S_{\triangle A B C}=\frac{\sqrt{3}}{4} \times 6^{2}=9 \sqrt{3}$, then
$V_{\text {=three-dimensional } D-A B C}=\frac{1}{3} D H \cdot S_{\triangle A B C}=3 \sqrt{39}$