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From the given system, it follows that $$(6x^2-3y^2-z-2z^2)-3(x^2-y^2-2z+1)-(6x-y+z^2-3)=0.$$ This equation can be rewritten as $(x-z)(x+z-2)=0$. Hence, $x=z$ or $x+z=2$.

If $x=z$. The given system is equivalent to $$\{\begin{array}{l}{x}^2+6x-y=3,\\ x^2-2x-y^2=-1.\end{array}$$ From the second equation, it obtains that $y^2=(x-1{)}^2$ or $y=\pm (x-1)$. Put $y=1-x$ and $1+x$ into the first equation, it obtains there are 4 solutions $(x,y,z)$ in this case, which are $$\left(\begin{array}{ccc}\frac{-5\pm \sqrt{33}}{2},& \frac{-7\pm \sqrt{33}}{2},& \frac{-5\pm \sqrt{33}}{2}\\ \frac{-7\pm \sqrt{65}}{2},& \frac{9\mp \sqrt{65}}{2},& \frac{-7\pm \sqrt{65}}{2}\end{array}\right).$$ If $x+z=2$. Letting $z=2-x$, the given system is equivalent to the following $$\{\begin{array}{l}{x}^2+2x+1-y=0,\\ x^2+2x-y^2-3=0.\end{array}$$ It follows that $1-y=-y^2-3$ or $y^2-y+4=0$. But this equation has no real roots, hence there is no solution for the given system in this case. ​