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.

$x^2+4y^2+16z^2=48 \cdots Eq.(1),\;\; xy+4yz+2zx=24 \cdots Eq.(2)$
From (1), $x^2+(2y)^2+(4z)^2=48 \cdots Eq.(3)$
$(2)\times4 \implies 4xy+16zy+8zx=96\cdots Eq.(4) $
$2\times Eq.(3)-Eq.(4)\implies 2[x^2+(2y)^2+(4z)^2]-2[2xy+8zy+4zx]=0 \implies (x-2y)^2+(2y-4z)^2+(x-4z)^2=0
\implies x=2y=4z.$
Now,
$xy+4zy+2zx=24 \implies 3x^2=48\implies x^2=16
$.
Now, $x=4$, then $y=2,z=1$.
$\implies x^2+y^2+z^2=21$.
You will get the same answer even if you take $x=-4$.
