<s>[INST] Find $x,y,z$ satisfying $x(y+z-x)=68-2x^2$, $y(z+x-y)=102-2y^2$, $z(x+y-z)=119-2z^2$ Solve for $x,y,z$:
$$x(y+z-x)=68-2x^2$$
$$y(z+x-y)=102-2y^2$$
$$z(x+y-z)=119-2z^2$$
After some manipulation, I obtain
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
After combining equations, I get
$$y=\frac{-51-x^2+z^2}{x-z}$$
This seems too tedious. Is there a simpler way?
 [/INST] Your system is equivalent to
\begin{align}
x(y+z+x)&=68\\
y(z+x+y)&=102\\
z(x+y+z)&=119.
\end{align}
Therefore, in effect, you have
\begin{align}
xa&=68\\
ya&=102\\
za&=119\\
a&=x+y+z.
\end{align}
This means that
$$a=\frac{68}{a}+\frac{102}{a}+\frac{119}{a}$$
or $$a^2=68+102+119\implies a=\pm 17,$$
which results in either $x= 4, y=6, z=7$ or $x=-4, y=-6, z=-7.$ (Thanks to paw88789 for pointing this out).
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