Five, (20 points)
Solve the system of equations: $x, y, z \in \mathbf{R}^{+}$
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+\sqrt{3} x y=20+8 \sqrt{3} \\
y^{2}+z^{2}=13 \\
z^{2}+x^{2}+x z=37
\end{array}\right.
$$

Find $(x, y, z)$.
Five. (20 points)
Take a point $O$ on the plane.
Draw $O A, O B, O C$,
such that $\angle A O B=150^{\circ}, \angle B O C$
$=90^{\circ}, \angle C O A=120^{\circ}$;
Let $O A=x, O B=y, O C=$
$z$,
then $A B=\sqrt{20+8 \sqrt{3}} \quad B C$
$=\sqrt{13} \quad A C=\sqrt{37}$
It is known that $(x, y, z)=(4,2,3)$ is a set of solutions that satisfy the conditions (which can be verified),
If there are other solutions $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$.
If $x^{\prime}>4$ then $y^{\prime}<2 . z^{\prime}>3$ then $x^{\prime}<4$ contradiction.
If $x^{\prime}<4$ similarly leads to a contradiction.
Therefore, the only solution that satisfies the conditions is $(4,2,3)$.