11. 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 $\qquad$
$$
\frac{\sqrt{2}}{24} \pi a^{3}
$$
11.【Analysis and Solution】The height of the regular tetrahedron $A O_{1}=\frac{\sqrt{6}}{3} a$, the center of the sphere must be at the center of the regular tetrahedron, i.e., at point $O$ on $A O_{1}$.

By the method of division (i.e., connecting point $O$ with $A, B, C, D$ to form four small tetrahedrons, and calculating the volume), we know that $O O_{1}=\frac{1}{4} A O_{1}=\frac{\sqrt{6}}{12} a$, thus $r=O E=$ $\sqrt{O O_{1}^{2}+O_{1} E^{2}}=\frac{\sqrt{2}}{4} a, \therefore V=\frac{4}{3} \pi r^{3}=\frac{\sqrt{2}}{24} \pi a^{3}$.