12. 2. $17^{\star \star}$ In $\triangle A B C$, $\angle C=90^{\circ}, \angle B=30^{\circ}, A C=2, M$ is the midpoint of $A B$. If $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$, the volume of the tetrahedron $A-B C M$ is . $\qquad$
This is a problem of figure folding in solid geometry -- a spatial figure is formed by folding a plane figure. To solve such problems, the properties of the original figure should be fully utilized. Given that $A C=$ $M A=M B=M C=2, B C=4 \cos 30^{\circ}=2 \sqrt{3}$, $A B=2 \sqrt{2}$, so $B C^{2}=4 \cdot 3=2^{2}+(2 \sqrt{2})^{2}=$ $A C^{2}+A B^{2}$. Therefore, the projection $H$ of $M$ on the plane $A B C$ is the circumcenter of the right triangle $\triangle A B C$, which is the midpoint of $B C$, as shown in the figure. In the isosceles triangle $\triangle B C M$, $M H=1$, thus $V_{A B C M}=V_{M-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot M H=$ $\frac{1}{3} \cdot 2 \sqrt{2} \cdot 1=\frac{2 \sqrt{2}}{3}$.