<s>[INST] If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$. 
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.

I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$
But I can't progress after this.
 [/INST] $$0 = x^2 + 4 y^2 + 16 z^2 - 2 (4yz + 2zx + xy) = (x-y-2z)^2 + 3 (y-2z)^2  $$
So, as noted $y = 2z$  and $x = y+2z = 4z,$ so
$$ 48 = x^2 + 4 y^2 + 16 z^2 = 16 z^2 + 16 z^2 + 16 z^2 = 48 z^2 $$
and $z = \pm 1.$
</s>