10. In the tetrahedron $P-ABC$, the three edges $PA$, $PB$, and $PC$ are pairwise perpendicular, and $PA=1$, $PB=PC=2$. If $Q$ is any point on the surface of the circumscribed sphere of the tetrahedron $P-ABC$, then the maximum distance from $Q$ to the plane $ABC$ is
10. $\frac{3}{2}+\frac{\sqrt{6}}{6}$.

Notice that, the circumscribed sphere of the tetrahedron $P-ABC$ is the circumscribed sphere of the rectangular parallelepiped with $PA$, $PB$, and $PC$ as its length, width, and height, respectively. Its diameter is $2R=\sqrt{1^{2}+2^{2}+2^{2}}=3$.
Also, $\cos \angle BAC=\frac{1}{5}$, so $\sin \angle BAC=\frac{2\sqrt{6}}{5}$.
Thus, the radius of the circumcircle of $\triangle ABC$ is
$$
r=\frac{BC}{2 \sin \angle BAC}=\frac{5\sqrt{3}}{6}.
$$

Therefore, the distance from the center of the sphere $O$ to the plane $ABC$ is
$$
\sqrt{R^{2}-r^{2}}=\frac{\sqrt{6}}{6}.
$$

Hence, the maximum distance from point $Q$ to the plane $ABC$ is $\frac{3}{2}+\frac{\sqrt{6}}{6}$.