2. Given a plane in space and a cube, the angle between the plane and each of the 12 edges of the cube is equal to $\alpha$, find the value of $\alpha$.


2. The difficulty of the problem lies in the fact that it is not easy to determine the position of the plane relative to the cube. From the given conditions, if we set the edges $AB, AC, AD$ of the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ to have the same angle with the given plane, we can conclude that the given plane is parallel to the face $BCD$. Furthermore, the face $BCD$ forms the same angle with all 12 edges of the cube, so we only need to find the angle formed by the edge $AD$ and the face $BCD$. For this purpose, draw $AH \perp$ face $BCD$, with $H$ being the projection of $A$ on the face $BCD$, and connect $DH$, so $\angle ADH=\alpha$. Noting that $\triangle BCD$ is an equilateral triangle, we can prove that $H$ is the circumcenter and centroid of $\triangle BCD$. If the edge length of the cube is $a$, then $DH=\frac{2}{3} \cdot CD \cdot \sin 60^{\circ}=\frac{\sqrt{6}}{3} a$, and $\angle AHD=90^{\circ}$, thus $\cos \alpha=\cos \angle ADH=\frac{DH}{AD}=\frac{\sqrt{6}}{3}$, hence $\alpha=\arccos \frac{\sqrt{6}}{3}$.