3. Given a tetrahedron $P-ABC$ with edge lengths $PA=1, PB=2, PC=3$, and $PA \perp PB, PB \perp PC, PC \perp PA$. Then the maximum distance from a moving point $Q$ on the surface of the circumscribed sphere of this tetrahedron to the plane $ABC$ is
3. $\frac{3}{7}+\frac{\sqrt{14}}{2}$.

Notice that, the circumsphere of the tetrahedron $P-ABC$ is the same as the circumsphere of the rectangular parallelepiped with $PA$, $PB$, and $PC$ as its length, width, and height, respectively. The diameter of this sphere is $2R=\sqrt{14}$.

Also, $\cos \angle BAC=\frac{\sqrt{2}}{10} \Rightarrow \sin \angle BAC=\frac{7\sqrt{2}}{10}$, thus, the radius of the circumcircle of $\triangle ABC$ is
$$
\begin{array}{l}
2r=\frac{BC}{\sin \angle BAC}=\frac{\sqrt{13}}{\frac{7\sqrt{2}}{10}}=\frac{5\sqrt{26}}{7} \\
\Rightarrow r=\frac{5\sqrt{26}}{14}.
\end{array}
$$

Therefore, the distance from the center $O$ of the sphere to the plane $ABC$ is
$$
\sqrt{R^2-r^2}=\sqrt{\left(\frac{\sqrt{14}}{2}\right)^2-\left(\frac{5\sqrt{26}}{14}\right)^2}=\frac{3}{7}.
$$

Thus, the maximum distance from point $Q$ to the plane $ABC$ is
$$
\frac{3}{7}+\frac{\sqrt{14}}{2}.
$$