9. Given $x, y, z \in \mathbf{R}_{+}$, and
$$
\begin{array}{l}
x^{2}+y^{2}+x y=1, y^{2}+z^{2}+y z=2, \\
z^{2}+x^{2}+z x=3 .
\end{array}
$$

Then $x+y+z=$
9. $\sqrt{3+\sqrt{6}}$.

As shown in Figure 3, construct a right-angled triangle
$$
\begin{array}{l}
\triangle A B C, A B=\sqrt{3}, C A= \\
\sqrt{2}, B C=1 .
\end{array}
$$

Let $F$ be the Fermat point of $\triangle A B C$, which satisfies
$$
\begin{array}{l}
\angle A F B=\angle B F C \\
=\angle C F A=\frac{2 \pi}{3} . \\
\text { Let } B F=x, \\
C F=y, A F=z .
\end{array}
$$

By the Law of Cosines, $x, y, z$ satisfy the three equations given in the problem, then $x+y+z$ is the sum of the distances from the Fermat point to each vertex.

Construct the circumcircle $\Gamma$ of $\triangle A F C$, connect $B F$ and extend it to intersect the circle $\Gamma$ at point $D$, connect $D A$ and $D C$, and take point $E$ on $D F$ such that $F E=F A$, connect $A E$. Then $\triangle A E F$ and $\triangle A C D$ are both equilateral triangles. Therefore, $\triangle A F C$ rotates $60^{\circ}$ to become $\triangle A E D$.
$$
\begin{array}{l}
\text { Hence } \triangle A F C \cong \triangle A E D \\
\Rightarrow F C=E D \\
\Rightarrow B D=F A+F B+F C .
\end{array}
$$

In $\triangle B C D$, by the Law of Cosines, we get
$$
\begin{array}{l}
B D=\sqrt{C B^{2}+C D^{2}-2 C B \cdot C D \cos \frac{5 \pi}{6}} \\
=\sqrt{3+\sqrt{6}} .
\end{array}
$$

Therefore, $x+y+z=B D=\sqrt{3+\sqrt{6}}$.