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this is my solution $6x-y+z^2=3(1)$ $x^2-y^2-2z=-1(2)$ $6x^2-3y^2-y-2z^2=0(3)$ equivalent $6x-y=3-z^2$ $3x^2-3y^2=6z-3$ $6x^2-3y^2-y=2z^2$ (3)-(2)-(1) so we have $3x^2-6x=3z^2-6z$ equivalent $x=z$ or $x+z=2$ 1. x=z so (*) equivalent $6x-y+x^2=3(1)$ $(x-1)^2=y^2(2)$ 1') $x-1=y$ so (1) equivalent $x^2+5x-2=0$ equivalent x=$ \frac {-5 + 33^{1/2}} {2}$ or x=$ \frac {-5 - 33^{1/2}}{2}$ so we have $ x=\frac {-5 + 33^{1/2}} {2}$ ;$y=\frac {-7+33^{1/2}}{2}$; $z=\frac {-5 + 33^{1/2}} {2}$ or $x=\frac {-5 - 33^{1/2}} {2}$;$y=\frac {-7-33^{1/2}}{2}$;$z=\frac{-5-33^{1/2}}{2}$ répectively with $1-x=y$ we have $x=z=\frac {-7+65^{1/2}}{2}$;$y=\frac{9-65^{1/2}}{2}$ or $x=z=\frac {-7-65^{1/2}}{2}$;$y=\frac{9+65^{1/2}}{2}$ 2)x+z=2 so (*) equivalent $6x-y-(2-x)^2=3$ $x^2-y^2-2(2-x)=-1$ $6x^2-3y^2-y-2(2-x)^2=0$ equivalent $x^2-10x+7=-y$ $3x^2+6x-9=3y^2$ $4x^2+8x-4=3y^2+y$ hent $2x^2-8x+12=0$ hent this system of equation have no solutions

Solution: $\left(\frac{-5+\sqrt{33}}{2},\frac{-7+\sqrt{33}}{2},\frac{-5+\sqrt{33}}{2}\right)$ $\left(\frac{-5-\sqrt{33}}{2},\frac{-7-\sqrt{33}}{2},\frac{-5-\sqrt{33}}{2}\right)$ $\left(\frac{-7+\sqrt{65}}{2},\frac{9-\sqrt{65}}{2},\frac{-7+\sqrt{65}}{2}\right)$ $\left(\frac{-7-\sqrt{65}}{2},\frac{9+\sqrt{65}}{2},\frac{-7-\sqrt{65}}{2}\right)$