10. Given that the four vertices of the tetrahedron $P-ABC$ lie on the surface of sphere $O$, $PA=PB=PC$, $\triangle ABC$ is an equilateral triangle with side length 2, $E, F$ are the midpoints of $AC, BC$ respectively, $\angle EPF=60^{\circ}$, then the surface area of sphere $O$ is $\qquad$.
According to the problem, $P-ABC$ is a regular tetrahedron, so $\triangle PEF$ is an equilateral triangle. Therefore, from $PE=1=\frac{1}{2} AC \Rightarrow PA \perp PC$, and $PA=PC=\sqrt{2}$. Thus, the circumscribed sphere of the regular tetrahedron $P-ABC$ is the same as the circumscribed sphere of a cube with edges $PA, PB, PC$, i.e., $2R=\sqrt{3} \cdot \sqrt{2}=\sqrt{6}$. Therefore, the surface area of sphere $O$ is $6\pi$.