For (a), we add the three equations to get $x^2 + y^2 + z^2 -4x-4y-4z+3p=0 \iff (x-2)^2+(y-2)^2+(z-2)^2+3p-12=0$. Thus, there are no solutions for $p>4$ For $p=4, x=y=z=2$ is a solution. For (b), we will show first that $x,y,z \ge 0$. Assume WLOG that $x<0$. Then $z^2+p-y=3x<0$ and $y^2-3z+p=x<0$. Since $y>z^2+p$ and $3z>y^2+p$ and $p>0$, we get $y,z>0$. From $p>1$, note that $z^2-y+p<0 \le (z-\sqrt{p})^2$ which implies $y>2z \sqrt{p}>2z$ and $y^2-3z+p<0 \le (y - \sqrt{p})^2$ which implies $z>\frac{2}{3}y \sqrt{p}> \frac{2}{3}y$. However, we get $y>2z>\frac{4}{3}y$, contradiction. Thus, $x,y,z \ge 0$. Now assume WLOG that $max\{x,y,z\}=x$. Then $z=x^2-3y+p \ge z^2- 3x+p=y$. So $x \ge z \ge y$ and $z=x^2-3y+p \ge y^2 -3z+p=x$. Thus, $x=z$. After that, we get $x^2-3y+p=z=x$ and $x^2-3x+p=z^2-3x+p=y$. Subtracting the two equations, we have $3(x-y)=x-y \iff x=y$. Thus, we conclude that $x=y=z$.