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.
 A: 
Please note that
$ (\sqrt3 a)^2 + b^2 - 2 (\sqrt3 a) b \cos 150^0 = 3a^2 + b^2 + 3ab = 75$
$b^2 + (\sqrt3 c)^2 - 2 b (\sqrt3 c) \cos 90^0 = b^2 + 3c^2 = 27$
$ (\sqrt3 a)^2 + (\sqrt3 c)^2 - 2 (\sqrt3 a) (\sqrt3 c) \cos 120^0 = 3a^2 + 3 c^2 + 3 a c = 48$
Angles add to $360^0$ so there is a point $O$ inside $\triangle PQR$ with $OP = \sqrt3 a, OQ = b, OR = \sqrt3 c$ and $PQ = \sqrt{75}, QR = \sqrt{27}$ and $PR = \sqrt{48}$
Next observe that $PQ^2 = QR^2 + PR^2$ which means $\triangle PQR$ is a right triangle.
$ \displaystyle S_{\triangle PQR} = \frac{1}{2} \cdot \sqrt{27} \cdot \sqrt{48} = 18$
But $\displaystyle S_{\triangle PQR} = S_{\triangle POR} + S_{\triangle QOR} + S_{\triangle POQ}$
As we know area of a triangle is $\frac{1}{2} a  b \sin \theta$ where $a, b$ are two sides with angle between them being $\theta$.
Adding individual areas we get to,
$\frac{\sqrt3}{4} (ab + 2bc + 3 ac) = 18$
So, $ab + 2bc+3ac = 24 \sqrt3$
A: 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
$a=-\sqrt{\frac{93184}{6553}-\frac{24576\sqrt{3}}{6553}}$,
$b=-\sqrt{\frac{58995}{6553}+\frac{31104\sqrt{3}}{6553}}$,
$c=-\sqrt{\frac{-10368\sqrt{3}}{6553}+\frac{39312}{6553}}$.
