2. (20 points) Given that $x$, $y$, and $z$ 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 the value of $x+y+z$.
2. The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-2 x y \cos 120^{\circ}=(\sqrt{3})^{2}, \\
y^{2}+z^{2}-2 y z \cos 120^{\circ}=2^{2}, \\
z^{2}+x^{2}-2 z x \cos 120^{\circ}=(\sqrt{7})^{2} .
\end{array}\right.
$$

This suggests the Law of Cosines, and we can construct a right triangle $\triangle ABC$, where $AB=\sqrt{3}, BC=2, AC=\sqrt{7}$.

Within the right triangle $\triangle ABC$, there exists a point $O$ which is the intersection of arcs with $120^{\circ}$ angles subtended by the chords $AB$ and $BC$, such that
$$
\angle AOB=\angle BOC=\angle COA=120^{\circ}.
$$

Let $OA=x, OB=y, OC=z$.
$$
\begin{array}{l}
\text { By } S_{\triangle AOB}+S_{\triangle BOC}+S_{\triangle COA}=S_{\triangle ABC} \\
\Rightarrow \frac{\sqrt{3}}{4} x y+\frac{\sqrt{3}}{4} y z+\frac{\sqrt{3}}{4} z x=\sqrt{3} \\
\Rightarrow x y+y z+z x=4 .
\end{array}
$$

Adding the three equations in (1) yields
$$
2\left(x^{2}+y^{2}+z^{2}\right)+x y+y z+z x=14 \text {. }
$$

Substituting equation (2) into this, we get
$$
x^{2}+y^{2}+z^{2}=5 \text {. }
$$

Therefore, $x+y+z$
$$
\begin{array}{l}
=\sqrt{x^{2}+y^{2}+z^{2}+2(x y+y z+z x)} \\
=\sqrt{13} .
\end{array}
$$