If the real numbers a, b, care such that $a^2+4b^2+16c^2=48$ and $ab+4bc+2ca=24$, then what is the value of ${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2$?

$a^2+4b^2+16c^2=48$

$2{\mathrm{a}}^2+8{\mathrm{b}}^2+32{\mathrm{c}}^2=96 .....\left(i\right)$

$4ab+16bc+8ac=96 .....\left(ii\right)$

From equation (i) and (ii), we get

$\begin{array}{l}(a-2b{)}^2+(2b-4c{)}^2+(4c-a{)}^2=0\\ \Rightarrow a=2b=4c\end{array}$

Then from equation (i)

$4b^2+4b^2+4b^2=48$

$\begin{array}{l}\Rightarrow 12b^2=48\\ \Rightarrow b=2\\ \therefore a=4,c=1\\ a^2+b^2+c^2=16+4+1=21\end{array}$