12. 3.6 $\star$ A sphere is tangent to all six edges of a regular tetrahedron. If the edge length of the tetrahedron is $a$, then the volume of the sphere is $\qquad$
As shown in the figure, let the center of the sphere be $O$, the radius be $r$, the volume of the sphere be $V$, the center of face $BCD$ be $O_{1}$, and the midpoint of edge $BC$ be $E$. Then $AO_{1} = \sqrt{a^{2} - O_{1}B^{2}} - \sqrt{a^{2} - \frac{1}{3}a^{2}} = \frac{\sqrt{6}}{3}a$. From $OB^{2} = O_{1}O^{2} + O_{1}B^{2} = (O_{1}A - OB)^{2} + O_{1}B^{2}$, we get $\frac{2}{3}a^{2} - \frac{2\sqrt{6}}{3}a \cdot OB + \frac{1}{3}a^{2} = 0$, hence $OB = \frac{\sqrt{6}}{4}a$. Therefore, $r = OE = \sqrt{OB^{2} - BE^{2}} = \sqrt{\frac{3}{8}a^{2} - \frac{1}{4}a^{2}} = \frac{1}{2\sqrt{2}}a$, so $V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi \cdot \frac{1}{16\sqrt{2}}a^{3} = \frac{\sqrt{2}}{24}\pi a^{3}$.