Consider a triangle $A B C$ with Fermat point $P$ such that $A P=a, B P=b, C P=c$. Then $$A B^{2}=A P^{2}+B P^{2}-2 A P \cdot B P \cos \left(120^{\circ}\right)$$ by the Law of Cosines, which becomes $$A B^{2}=a^{2}+a b+b^{2}$$ and hence $A B=3$. Similarly, $B C=\sqrt{52}$ and $A C=7$. Furthermore, we have $$\begin{aligned} B C^{2}=52 & =A B^{2}+B C^{2}-2 A B \cdot B C \cos \angle B A C \\ & =3^{2}+7^{2}-2 \cdot 3 \cdot 7 \cos \angle B A C \\ & =58-42 \cos \angle B A C \end{aligned}$$ And so $\cos \angle B A C=\frac{1}{7}$. Invert about $A$ with arbitrary radius $r$. Let $B^{\prime}, P^{\prime}, C^{\prime}$ be the images of $B, P, C$ respectively. Since $\angle A P B=\angle A B^{\prime} P^{\prime}=120^{\circ}$ and $\angle A P C=\angle A C^{\prime} P^{\prime}=120^{\circ}$, we note that $\angle B^{\prime} P^{\prime} C^{\prime}=120^{\circ}-\angle B A C$, and so $$\begin{aligned} \cos \angle B^{\prime} P^{\prime} C^{\prime} & =\cos \left(120^{\circ}-\angle B A C\right) \\ & =\cos 120^{\circ} \cos \angle B A C-\sin 120^{\circ} \sin \angle B A C \\ & =-\frac{1}{2}\left(\frac{1}{7}\right)+\frac{\sqrt{3}}{2}\left(\frac{4 \sqrt{3}}{7}\right) \\ & =\frac{11}{14} \end{aligned}$$ Furthermore, using the well-known result $$B^{\prime} C^{\prime}=\frac{r^{2} B C}{A B \cdot A C}$$ for an inversion about $A$, we have $$\begin{aligned} B^{\prime} P^{\prime} & =\frac{B P r^{2}}{A B \cdot A P} \\ & =\frac{b r^{2}}{a \cdot 3} \\ & =\frac{b r^{2}}{3 a} \end{aligned}$$ and similarly $P^{\prime} C^{\prime}=\frac{c r^{2}}{7 a}, B^{\prime} C^{\prime}=\frac{r^{2} \sqrt{52}}{21}$. Applying the Law of Cosines to $B^{\prime} P^{\prime} C^{\prime}$ gives us $$\begin{aligned} B^{\prime} C^{\prime 2} & =B^{\prime} P^{\prime 2}+P^{\prime} C^{2}-2 B^{\prime} P^{\prime} \cdot P^{\prime} C^{\prime} \cos \left(120^{\circ}-\angle B A C\right) \\ \Longrightarrow \frac{52 r^{4}}{21^{2}} & =\frac{b^{2} r^{4}}{9 a^{2}}+\frac{c^{2} r^{4}}{49 a^{2}}-\frac{11 b c r^{4}}{147 a^{2}} \\ \Longrightarrow \frac{52}{21^{2}} & =\frac{b^{2}}{9 a^{2}}+\frac{c^{2}}{49 a^{2}}-\frac{11 b c}{147 a^{2}} \\ \Longrightarrow \frac{52}{21^{2}} & =\frac{49 b^{2}-33 b c+9 c^{2}}{21^{2} a^{2}} \end{aligned}$$ and so $\frac{49 b^{2}-33 b c+9 c^{2}}{a^{2}}=52$. Motivation: the desired sum looks suspiciously like the result of some Law of Cosines, so we should try building a triangle with sides $\frac{7 b}{a}$ and $\frac{3 c}{a}$. Getting the $-\frac{33 b c}{a}$ term is then a matter of setting $\cos \theta=\frac{11}{14}$. Now there are two possible leaps: noticing that $\cos \theta=\cos (120-\angle B A C)$, or realizing that it's pretty difficult to contrive a side of $\frac{7 b}{a}$ - but it's much easier to contrive a side of $\frac{b}{3 a}$. Either way leads to the natural inversion idea, and the rest is a matter of computation.