$x=\sqrt{64-3z+4y-y^{2}}+8$
$x=-\sqrt{64-3z+4y-y^{2}}+8$

$y=\sqrt{4-3z+16x-x^{2}}+2$
$y=-\sqrt{4-3z+16x-x^{2}}+2$

$x=\sqrt{64-3z+4y-y^{2}}+8$
$x=-\sqrt{64-3z+4y-y^{2}}+8\text{, }z\leq \frac{64+4y-y^{2}}{3}$

$y=\sqrt{4-3z+16x-x^{2}}+2$
$y=-\sqrt{4-3z+16x-x^{2}}+2\text{, }z\leq \frac{4+16x-x^{2}}{3}$