Question 150: Four spheres with a radius of 1 are pairwise externally tangent. Then the volume of the circumscribed regular tetrahedron is $\qquad$ .
For a regular tetrahedron with edge length $\mathrm{a}$, let the center of the base $\triangle B C D$ be $P$, and the center of the circumscribed sphere be $O$. It is easy to know that the height $h=\frac{\sqrt{6} a}{3}$, which also means $a=\frac{\sqrt{3} h}{\sqrt{2}}$. Therefore, its volume is $\frac{1}{3} \cdot \frac{\sqrt{3} a^{2}}{4} \cdot h=\frac{\sqrt{3} h^{3}}{8}$. Furthermore, $A O=\frac{3 \mathrm{~h}}{4}=\frac{\sqrt{6} \mathrm{a}}{4}, P O=\frac{\mathrm{h}}{4}=\frac{\sqrt{6 \mathrm{a}}}{12}$.

Four pairwise externally tangent spheres with a radius of 1 form a regular tetrahedron with an edge length of 2. According to the above formula, the height of the circumscribed regular tetrahedron is $\left(\frac{\sqrt{6}}{12} \times 2+1\right) \times 4=\frac{2(6+\sqrt{6})}{3}$. Therefore, its volume is $V=\frac{\sqrt{3}\left(\frac{2(6+\sqrt{6}}{3}\right)^{3}}{8}=$ $\frac{\sqrt{3}(6+\sqrt{6})^{3}}{27}$.