5. Given a plane that makes an angle $\alpha$ with all 12 edges of a cube, then $\sin \alpha=$
$$
\frac{\sqrt{3}}{3}
$$
5.【Analysis and Solution】As shown in the figure, according to the problem, the plane $BCD$ forms an angle $\alpha$ with all 12 edges of the cube. Draw $AH \perp$ plane $BCD$ through point $A$, and connect $DH$, then $\alpha=\angle ADH$. Let the side length of the cube be $a$, then it is easy to find that $DH=\frac{2}{3} \times \sqrt{2} a \cdot \sin 60^{\circ}$
$$
\begin{array}{c}
=\frac{\sqrt{6}}{3} a, \\
AH=\sqrt{a^{2}-\left(\frac{\sqrt{6}}{3} a\right)^{2}}=\frac{\sqrt{3}}{3} a . \\
\therefore \sin \alpha=\sin \angle ADH=\frac{AH}{AD}=\frac{\sqrt{3}}{3} .
\end{array}
$$