3. The volume of the solid of revolution obtained by rotating a triangle with side lengths $3, 4, 5$ around the angle bisector of its largest interior angle is $\qquad$ .
3. $\frac{32 \sqrt{2} \pi}{7}$.

As shown in Figure 5, $CD$ is the bisector of $\angle ACB=90^{\circ}$, point $D$ is on $AB$, and the symmetric point of $A$ about $CD$ is $A_{1}$. $AA_{1}$ intersects the extension of $CD$ at point $E$. The volume of the solid of revolution is equal to the volume obtained by rotating $\triangle ACD$ around $CD$, which is
$$
\begin{array}{l}
V=\frac{1}{3} \pi \cdot AE^{2}(CE-DE) \\
=\frac{\pi}{3}\left(4 \sin 45^{\circ}\right)^{2} \cdot CD=\frac{8 \pi}{3} \times 4 \times \frac{\sin A}{\sin \left(A+45^{\circ}\right)} \\
=\frac{32 \pi}{3} \times \frac{\frac{3}{5}}{\frac{3}{5} \times \frac{1}{\sqrt{2}}+\frac{4}{5} \times \frac{1}{\sqrt{2}}}=\frac{32 \sqrt{2} \pi}{7} .
\end{array}
$$