3. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of side $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the distance from point $M$ to plane $A B C$ is $\qquad$
3. 1 .

From the problem, we know that $M A=M B=M C=2$.
Therefore, the projection $O$ of $M$ on the plane $A B C$ is the circumcenter of $\triangle A B C$.
Given $A B=2 \sqrt{2}, A C=2, B C=2 \sqrt{3}$, we know that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $O$ is the midpoint of the hypotenuse $B C$ of the right triangle $\triangle A B C$.
Hence, $M O=\sqrt{M C^{2}-O C^{2}}=1$ is the desired result.