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

$a^2+ab+b^2=9   \implies a^2 = 9 - ab -  b^2 $   
$b^2+bc+c^2=52 \implies b^2 = 52 -bc - c^2$
$c^2+ac+a^2=49 \implies c^2 = 49 - ac - a^2$
Substituting $a^2$ in $(iii)$   
$(c-b)(a+b+c) = 40 --(iv)$  
Similarly,  
$(b-a)(a+b+c) = 3 -- (v)$  
$(c-a)(a+b+c)= 43 -- (vi)$
Now adding $(i),(ii),(iii)$  
$2(a^2 + b^2 + c^2) + ab + bc + ca = 110$  
$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca + a^2 + b^2 + c^2- ab - bc - ca = 110$  
$(a+b+c)^2 + \cfrac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2} = 110$  
Now, substitute $(a-b),(b-c),(c-a)$ from equation $(iv),(v),(vi)$,  
$(a+b+c)^2 + \cfrac{3^2+40^2+ 43^2}{2(a+b+c)^2}= 110$  
Take,$(a+b+c)^2 = t$ and solving quadratic,  
$t = 91,19$  
$(a+b+c) = \sqrt{91}$ or $\sqrt{19}$  
clearly, $a+b+c$ can't be $\sqrt{19}$
Now, substitue $a+b+c$ in $(iv),(v),(vi)$  
we get, the values of (c-b),(b-a),(c-a)  
put the values of b,c in $a+b+c = \sqrt{91} $  
We will get the value of a,b,c.  
