9. Given $x, y, z \in \mathbf{R}$, and $x^{2}+y^{2}+x y=1, y^{2}+z^{2}+y z=2, x^{2}+z^{2}+x z=3$, then $x+y+z=$
Given $\angle A P B=\angle B P C=\angle C P A=120^{\circ}$,
$P A=x, P B=y, P C=z$, then $A B=1$,
$$
B C=\sqrt{2}, A C=\sqrt{3} \Rightarrow \angle A B C=90^{\circ} \text {. }
$$

Construct equilateral triangles $\triangle P_{1} P B$ and $\triangle A_{1} B C$,
connect $P P_{1}, P_{1} A_{1}$, by $\angle P B C=\angle P_{1} B A_{1}$
we get $\triangle P B C \sim \triangle P_{1} B A_{1} \Rightarrow A, P, P_{1}, A_{1}$
are collinear $\Rightarrow A A_{1}=x+y+z$.
In $\triangle A B A_{1}$, $A B=1, B A_{1}=\sqrt{2}, \angle A B A_{1}=150^{\circ}$,
so $A A_{1}^{2}=1+2+2 \cdot 1 \cdot \sqrt{2} \cdot \frac{\sqrt{3}}{2}=3+\sqrt{6} \Rightarrow x+y+z=\sqrt{3+\sqrt{6}}$.