Answer

The first equation of the system is written: $(x^{2}-6xz+9z^{2})+(y^{2}-8yz+16z^{2})=0\Rightarrow (x-3z)^{2}+(y-4z)^{2}=0$. Note that $(x-3z)^{2}+(y-4z)^{2} \ge 0$, where the equality holds when $(x-3z)^{2}=0, (y-4z)^{2}=0 \Rightarrow x-3z=0, y-4z=0 \Rightarrow x=3z, y=4z$. The second equation of the system is now written: $27z^{2}+32z^{2}+z^{2}=60z^{2}=240 \Rightarrow z^{2}=4 \Rightarrow z=-2, +2$. Hence the triplets: $(x,y,z)=(6,8,2),(-6,-8,-2)$.