Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$ 
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is

Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So we have
$OA=a,OB=b,OC=c$ and using cosine formula we have $AB=3,BC=7,CA=\sqrt{52}$.
Now using Area of triangle is $\displaystyle ab+bc+ca=\frac{42}{\sqrt{3}}$.
Could some help me to solve it , Thanks

We can treat the equations as quadratic with variables $a,b,c$, and thus we solve for each of their values using the quadratic formula:
$$\begin{align}
a\to& \frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{1}\\
b\to& \frac{1}{2} \left(\sqrt{208-3 c^2}-c\right)\tag{2}\\
c\to& \frac{1}{2} \left(\sqrt{196-3 a^2}-a\right)\tag{3}\\
\end{align}$$
Since $a,b,c \in \mathbb{R}^+$, then we can discard the negative values of $a,b,c$.  We can rewrite $(3)$ in terms of $a$, and we get:
$$a\to \frac{1}{2} \left(\sqrt{196-3 c^2}-c\right)\tag{4}$$
Then we substitute $(4)$ in $(1)$ and we get:
$$\frac{1}{2} \left(\sqrt{196-3 c^2}-c\right)=\frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{5}$$
Using $(2)$ and substituting it in $(5)$, we have:
$$2 \sqrt{196-3 c^2}+\sqrt{208-3 c^2}=\sqrt{6} \sqrt{c^2+\sqrt{208-3 c^2} c-80}+3 c$$
For which we get $c=\frac{58}{\sqrt{91}}$, using that we get $a$ and $b$ from $(4)$ and $(2)$, respectively:
$$\begin{align}
a=&\frac{15}{\sqrt{91}}\\
b=&\frac{18}{\sqrt{91}}\\
\end{align}$$
And thus, we can now solve the problem:
$$\begin{align}\displaystyle \frac{49b^2-33bc+9c^2}{a^2}\Rightarrow &\displaystyle \frac{49(\frac{18}{\sqrt{91}})^2-33(\frac{18}{\sqrt{91}})(\frac{58}{\sqrt{91}})+9(\frac{58}{\sqrt{91}})^2}{(\frac{15}{\sqrt{91}})^2}\\
&\Rightarrow 52
\end{align}$$
