3. In the tetrahedron $S-ABC$, $BC=3, AC=4$, the distance from point $C$ to the midpoint of $AB$ is $\frac{5}{2}$, and the dihedral angles between the lateral faces $SAB$, $SBC$, and $SAC$ and the base are all $60^{\circ}$. Then the volume of the tetrahedron is $\qquad$
3. $2 \sqrt{3}$.

Since the angles between the three lateral faces and the base are equal, the projection of vertex $S$ onto the base is the incenter of $\triangle A B C$.

From the given information, $\angle A C B=90^{\circ}$, and $A B=5$. Therefore, the inradius $r$ of $\triangle A B C$ is $1$. Thus, the distance $h$ from point $S$ to the base $A B C$ is $\sqrt{3}$.
Hence, the required volume is $\frac{1}{3} S_{\triangle A B C} h=2 \sqrt{3}$.