3. $\frac{\sqrt{6}}{2}$.

Let the projection of point $P$ on plane $\alpha$ be $N$, then the length of line segment $P N$ is the distance from point $P$ to plane $\alpha$. Let $P N=h$, then $N A=\sqrt{3} h, N M=h, N C=\frac{h}{\sqrt{3}}$. In $\triangle N A B$, by the median length formula, we have
$$
4 N M^{2}+A B^{2}=2\left(N A^{2}+N B^{2}\right),
$$

which is
$$
4 h^{2}+2^{2}=2\left(3 h^{2}+\frac{h^{2}}{3}\right)
$$

Solving for $h$ gives $h=\frac{\sqrt{6}}{2}$.
3. Given that point $M$ is the midpoint of line segment $A B$ in plane $\alpha$, and point $P$ is a point outside plane $\alpha$. If $A B=2$, and the angles formed by lines $P A, P M, P B$ with plane $\alpha$ are $30^{\circ}, 45^{\circ}, 60^{\circ}$ respectively, then the distance from point $P$ to plane $\alpha$ is $\qquad$ (Provided by Li Hong)