Given a tetrahedron $P A B C$ with the three dihedral angles at the apex all being $60^{\circ}$, and the areas of the three lateral faces being $\frac{\sqrt{3}}{2}, 2, 1$, then the volume of this tetrahedron is $\qquad$ .
10. $\frac{2 \sqrt{6}}{9}$.

If the three lateral edges PA, PB, PC have lengths $a, b, c$ respectively, then
$$
\left\{\begin{array} { l } 
{ \frac { 1 } { 2 } a b \operatorname { s i n } 6 0 ^ { \circ } = \frac { \sqrt { 3 } } { 2 } , } \\
{ \frac { 1 } { 2 } b x \operatorname { s i n } 6 0 ^ { \circ } = 2 , } \\
{ \frac { 1 } { 2 } c a \operatorname { s i n } 6 0 ^ { \circ } = 1 }
\end{array} \text { solving gives } \left\{\begin{array}{l}
a=1, \\
b=2, \\
c=\frac{4}{3} \sqrt{3} .
\end{array}\right.\right.
$$

Taking $P C^{\prime}=P B^{\prime}=1$ on $P C^{\prime}, P B$ respectively, then $P-\Lambda B^{\prime} C^{\prime}$ is a regular tetrahedron with edge length 1, and its height is $h$ $=\frac{\sqrt{6}}{3}$. That is, the distance from point $A$ to the plane $P B C$ is $\frac{\sqrt{6}}{3}$, so $V_{\triangle A B C}=\frac{1}{3} S_{\triangle P B C} \cdot h=\frac{2 \sqrt{6}}{9}$.