10. 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 $\qquad$
Answer $\sqrt{3}$.
Analysis: Restore the tetrahedron into a rectangular box, and let the length, width, and height of the box be $a, b, c$, respectively, then
$$
\left\{\begin{array}{l}
a^{2}+b^{2}=10 \\
b^{2}+c^{2}=8 \\
a^{2}+c^{2}=6
\end{array} \quad \Rightarrow a^{2}+b^{2}+c^{2}=12\right.
$$

Therefore, the radius of the circumscribed sphere of the tetrahedron is $\sqrt{3}$.