Three, (20 points) Solve the system of equations
$$
\left\{\begin{array}{l}
x(y+z-x)=39 \cdots 2 x^{2}, \\
y(z+x-y)=52-2 y^{2}, \\
(x+y-z)=78-2 z^{2} .
\end{array}\right.
$$
$$
\begin{array}{l}
\text { Three, (1)+(2)+(3) gives } \\
(x+y+z)^{2}=169 .
\end{array}
$$

Taking the square root gives $x+y+z= \pm 13$,
Dividing (1), (2), (3) by (1) gives
$$
(x, y, z)=(3,4,6), \quad(-3,-4,-6) .
$$