12.2.13 ** Given that a plane forms an angle $\alpha$ with all 12 edges of a cube, then $\sin \alpha=$ $\qquad$
A cube, despite having 12 edges, has only three different "directions" for all its edges, so we only need to find a plane that forms equal angles with the three edges meeting at a vertex. As shown in Figure a, it is easy to see that plane BCD forms an angle of $\alpha$ with all 12 edges of the cube. Draw $AH \perp$ plane BCD, with H being the foot of the perpendicular, and connect DH, then $\alpha=\angle ADH$. Let the edge length of the cube be $b$, then $DH=\frac{2}{3} \cdot \sqrt{2} b \cdot \sin 60^{\circ}=\frac{\sqrt{6}}{3} b$, $AH=\sqrt{b^{2}-\left(\frac{\sqrt{6}}{3} b\right)^{2}}=\frac{\sqrt{3}}{3} b$, hence $\sin \alpha=\sin \angle ADH=\frac{AH}{AD}=\frac{\sqrt{3}}{3}$.
Figure a
Figure b