3. Given in the tetrahedron $S-ABC$, the sum of the three dihedral angles at each vertex of the base triangle is $180^{\circ}$, and the sides of the base triangle are $\sqrt{3}$, $2$, and $\sqrt{5}$. Then the volume of the tetrahedron is $\qquad$.
3. $\frac{\sqrt{6}}{3}$.

As shown in Figure 5, the three lateral faces of the tetrahedron $S-ABC$ are unfolded on the base plane.
$$
\begin{array}{r}
\because S_{1} A=S_{2} A \\
=S A, \text { and } \angle S_{1} A B+ \\
\angle B A C+\angle S_{2} A C=
\end{array}
$$
$180^{\circ}$,
$$
\therefore S_{1} 、 A 、 S_{2} \text { three }
$$

points are collinear, and $A$ is the midpoint of $S_{1} S_{2}$.
Similarly, $S_{1} 、 C 、 S_{3}$ are collinear, and $C$ is the midpoint of $S_{1} S_{3}$.
$\therefore$ In $\triangle S_{1} S_{2} S_{3}$, $A C=\frac{1}{2} S_{2} S_{3}=S B$.
Similarly, $A B=S C, B C=S A$.
Therefore, the three pairs of opposite edges of the tetrahedron $S-ABC$ are equal in length. Next, we construct a rectangular prism to find the volume of the tetrahedron. As shown in Figure 6, the length, width, and height of the rectangular prism are $x, y, z$, and the three face diagonals are $\sqrt{3}, 2$, and $\sqrt{5}$.

Then $\left\{\begin{array}{l}x^{2}+y^{2}=(\sqrt{3})^{2}, \\ y^{2}+z^{2}=(2)^{2}, \\ z^{2}+x^{2}=(\sqrt{5})^{2} .\end{array}\right.$ We get $\left\{\begin{array}{l}x=\sqrt{2}, \\ y=1, \\ z=\sqrt{3} .\end{array}\right.$
Thus, the volume of the tetrahedron
$$
\begin{aligned}
V & =V_{\text {rect }}-4 V_{D-S S B} \\
& =V_{\text {rect }}-4 \times \frac{1}{6} V_{\text {rect }}=\frac{1}{3} V_{\text {rect }}=\frac{\sqrt{6}}{3} .
\end{aligned}
$$