A cylinder can be placed inside a cube with an edge length of 1, such that the central axis of the cylinder coincides with one of the cube's body diagonals. Then the maximum volume of this cylinder is $\qquad$ .
12. $\frac{\sqrt{3}}{18} \pi$.

Let the radius of the base of the cylinder be $r$, and the height of the cylinder be $h$, then we have $\left(\frac{1}{2}(\sqrt{3}-h)\right)^{2}+r^{2}-\left(\frac{a}{2}\right)^{2}$. Here $\frac{\sqrt{3}}{2} a=$ $3 r$, thus $h=\sqrt{3}-2 \sqrt{2} r$. Therefore, the volume of the cylinder $V=\pi r^{2} h=\pi r^{2}(\sqrt{3}-2 \sqrt{2} r) \leqslant \frac{1}{2} \cdot \pi \cdot$ $\left(\frac{(\sqrt{3}-2 \sqrt{2} r)+\sqrt{2} r+\sqrt{2} r}{3}\right)^{3}=\frac{\sqrt{3}}{18} \pi$, equality holds when $r=\frac{\sqrt{6}}{6}$.