3. Given $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$, and the vector $\alpha=\left(\cos \frac{A-B}{2}, \sqrt{3} \sin \frac{A+B}{2}\right)$, $|\alpha|=\sqrt{2}$. If when $\angle C$ is maximized, there exists a moving point $M$ such that $|M A|, |A B|, |M B|$ form an arithmetic sequence, the maximum value of $|A B|$ is $\qquad$
3. $\frac{2 \sqrt{3}+\sqrt{2}}{4}$

Analysis: On one hand, from $|\alpha|=\sqrt{2}$, we know that $\cos ^{2} \frac{A-B}{2}+3 \sin ^{2} \frac{A+B}{2}=2$, which means $\cos (A-B)+3 \cos C=0$, so $\cos C=-\frac{\cos (A-B)}{3} \geq-\frac{1}{3}$. Therefore, when $\angle C$ is maximized, $\angle A=\angle B$, and $\cos C=-\frac{1}{3}$.
On the other hand, from $|M A| \backslash|A B| 、|M B|$ forming an arithmetic sequence, we know that point $M$ lies on the ellipse with foci at $A$ and $B$, and the length of the major axis is $2|A B|$. If we set the coordinates of $A$ and $B$ to be $(-1,0)$ and $(1,0)$, respectively, then the equation of the ellipse is: $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. From $\cos C=-\frac{1}{3}$, we get $C\left(0, \frac{\sqrt{2}}{2}\right)$. Let the coordinates of $M$ be $(2 \cos \theta, \sqrt{3} \sin \theta)$.