10. In the tetrahedron $P-ABC$, $PA=BC=\sqrt{6}$, $PB=AC=\sqrt{8}$, $PC=AB=\sqrt{10}$. Then the radius of the circumscribed sphere of the tetrahedron is $\qquad$.
10. $\sqrt{3}$.

To restore the tetrahedron into a rectangular box, let the length, width, and height of the box be $a$, $b$, and $c$ respectively. Then 
\[
\left\{\begin{array}{l}
a^{2}+b^{2}=10, \\ 
b^{2}+c^{2}=8, \\ 
a^{2}+c^{2}=6
\end{array} \Rightarrow a^{2}+b^{2}+c^{2}=12\right.
\]
Therefore, the radius of the circumscribed sphere of the tetrahedron is $\sqrt{3}$.