<s>[INST] Finding three unknowns from three equations Let $a$,$b$ and $c$ be three positive real numbers such that
$$\begin{cases}3a^2+3ab+b^2&=&75\\
b^2+3c^2&=&27\\c^2+ca+a^2&=&16\end{cases}$$
Find the value of  $ ab+2bc+3ca$.
My attempt: I observed that $3 . 16+27=75$. Then on replacing $16$ by $c^2+ca+a^2$,  $27$ by $b^2+3c^2$ and $75$ by $3a^2+3ab+b^2$, I got $2c^2+ca=ab$.
But after this I am unable to proceed.
Is there a way to proceed from here?
Any constructive hint is appreciated.
 [/INST] There are two values of the expression.
The value of
$ ab+2bc+3ca=-24\sqrt{3}$
and
$ ab+2bc+3ca=+24\sqrt{3}$.
The respective values of $a$, $b$ and $c$ are:
$a=+\sqrt{\frac{24576\sqrt{3}}{6553}+\frac{93184}{6553}}$,
$b=+\sqrt{\frac{58995}{6553}-\frac{31104\sqrt{3}}{6553}}$,
$c=-\sqrt{\frac{10368\sqrt{3}}{6553}+\frac{39312}{6553}}$;
and
$a=-\sqrt{\frac{93184}{6553}-\frac{24576\sqrt{3}}{6553}}$,
$b=-\sqrt{\frac{58995}{6553}+\frac{31104\sqrt{3}}{6553}}$,
$c=-\sqrt{\frac{-10368\sqrt{3}}{6553}+\frac{39312}{6553}}$.
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