12. 2. 3 ** As shown in Figure a, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$, respectively. The sine value of the plane angle formed by the dihedral angle between plane $C E B_{1}$ and plane $D_{1} F B_{1}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) 1
Parse as shown in figure b, extend $C E$, $D_{1} F$, and $D A$. It is easy to see that $C E$, $D_{1} F$, and $D A$ intersect at a point $G$. Connect $B_{1} G$. Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be 1. Then $B_{1} C=\sqrt{2}$, $B_{1} G=\sqrt{3}$, $C G=\sqrt{5}$, satisfying $B_{1} C^{2}+B_{1} G^{2}=C G^{2}$. Therefore, $B_{1} C \perp B_{1} G$. Similarly, $B_{1} D_{1} \perp B_{1} G$. Hence, $\angle C B_{1} D_{1}$ is the plane angle of the dihedral angle $C-B_{1} G-D_{1}$ formed by the planes $C E B_{1}$ and $D_{1} F B_{1}$.

Connect $C D_{1}$. In $\triangle B_{1} C D_{1}$, since $B_{1} C=B_{1} D_{1}=C D_{1}=\sqrt{2}$, therefore, $\angle C B_{1} D_{1}=$ $60^{\circ}$. Hence, the sine value of the plane angle of the dihedral angle formed by the planes $C E B_{1}$ and $D_{1} F B_{1}$ is $\frac{\sqrt{3}}{2}$. Therefore, the answer is C.团a
固 $b$