10. Given that $x, y, z$ are positive numbers, and satisfy $\left\{\begin{array}{l}x^{2}+y^{2}+x y=3, \\ y^{2}+z^{2}+y z=4, \\ z^{2}+x^{2}+z x=7,\end{array}\right.$ find $x+y+z$.
As shown in the figure, $P$ is the Fermat point of $\triangle A B C$. By rotating $\triangle P B C$ around point $B$ by $60^{\circ}$, we obtain $\triangle E B D$. Connecting $P E$, we find that $\triangle B P E$ is an equilateral triangle, thus $A, P, E, D$ are collinear $\Rightarrow A D=x+y+z$.
In $\triangle A B D$, $\angle A B D=90^{\circ}+60^{\circ}=150^{\circ}$, hence $A D^{2}=3+4+2 \cdot \sqrt{3} \cdot 2 \cdot \frac{\sqrt{3}}{2}=13$ $\Rightarrow A D=\sqrt{13}$, so $x+y+z=\sqrt{13}$.