5. Given that a plane forms an angle $\alpha$ with all 12 edges of a cube, then $\sin \alpha=$ $\qquad$
5. 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 $\alpha = \angle ADH$. Assuming the side length of the cube is $a$, it is easy to find that
$$
\begin{array}{c}
D H = \frac{2}{3} \times \sqrt{2} a \cdot \sin 60^{\circ} = \frac{\sqrt{6}}{3} a, \\
A H = \sqrt{a^{2} - \left(\frac{\sqrt{6}}{3} a\right)^{2}} = \frac{\sqrt{3}}{3} a . \\
\therefore \sin \alpha = \sin \angle ADH = \frac{A H}{A D} = \frac{\sqrt{3}}{3} .
\end{array}
$$