36th Hellenic Mathematical Olympiad

The first equation can be written in the form: $$\begin{array}{rlrl}{x}^2-6xz+9z^2+y^2-8yz+16z^2& =0& \iff (x-3z{)}^2+(y-4z{)}^2& =0\\ \iff x-3z& =0\text{and}y-4z& =0& \iff x=3z\text{and}y=4z.\end{array}$$ Therefore, all triples satisfying the first equation of the system are the following: $$(x,y,z)=(3t,4t,t),t\in \mathbb{R}.$$ Then the second equation takes the form: $$3\cdot 9t^2+2\cdot 16t^2+t^2=240\iff 60t^2=240\iff t^2=4\iff t=\pm 2.$$ For $t=2$, we get the solution $(x,y,z)=(6,8,2)$, while for $t=-2$ we get: $$(x,y,z)=(-6,-8,-2).$$