3. In the plane $\alpha$ there is a $\triangle A B C, \angle A B C=105^{\circ}$, $A C=2(\sqrt{6}+\sqrt{2})$. On both sides of the plane $\alpha$, there are points $S, T$, satisfying $S A=S B=S C=\sqrt{41}, T A=T B=T C=$ 5. Then $S T=$ $\qquad$.
3. 8 .

From Figure 4, since $S A=S B=S C$, we know that the projection $D$ of $S$ on plane $\alpha$ is the circumcenter of $\triangle A B C$. Similarly, the projection of $T$ on plane $\alpha$ is also the circumcenter of $\triangle A B C$. By the uniqueness of the circumcenter, we have $S T \perp a$.
Connect $A D$. By the Law of Sines, we have
T
$$
\begin{array}{l}
A D=\frac{A C}{\sin \angle A B \bar{C}} \\
=\frac{3(\sqrt{6}+\sqrt{2})}{2\left(\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}\right)}=4 .
\end{array}
$$

Thus,
$$
\begin{aligned}
S T & =S D+D T \\
& =\sqrt{S A^{2}-A D^{2}}+\sqrt{T A^{2}-A D^{2}} \\
& =\sqrt{41-16}+\sqrt{25-16}=5+3=8
\end{aligned}
$$