<s>[INST] Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$ 
Solve the following system of equations in $\Bbb R^+$:
   $$
\left\{ 
\begin{array}{l}
xy+yz+xz=12 \\ 
xyz=2+x+y+z\\ 
\end{array}
\right. 
$$  

I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
 [/INST] I will use the cubic equation with roots $x$, $y$, $z$, inspired by the answer of @dxiv:. 
$$X^3 - s X^2 + 12 X - (2+s) =0$$
The discriminant of the equation is $-(2s+13)(2s+15)(s-6)^2$. If $s\ge 0$ then the discriminant is positive only for $s=6$, which gives $a=b=c=2$. Therefore: the system has only one positive solution, but infinitely many negative ones, for each $s \in [-\frac{15}{2}, -\frac{13}{2}]$ we get a solution with $a+b+c = s$. For instance, for $s=-\frac{13}{2}$, we get $a=b=-3$, $c=-\frac{1}{2}$ a solution. 
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