Think beyond numbers.

The lengths $a,b,c$ are the distances $PX,PY,PZ$ from a point $P$ to the vertices of a triangle $XYZ$ which has sides $x=2\sqrt{13}$ , $y=7$ and $z=3$ , where angles $\angle XPY\,,\, \angle YPZ\,,\,\angle ZPX$ are all $120^\circ$ , and hence $P$ is the first Fermat point of the triangle $XYZ$ . The Cosine Rule tells us that $\cos X \; = \; \tfrac17 \hspace{1.5cm} \cos Y \; = \; \tfrac{1}{\sqrt{13}} \hspace{1.5cm} \cos Z \; = \; \tfrac{23}{7\sqrt{13}}$ so the triangle $XYZ$ is acute-angled.

We have (relative) trilinear coordinates $\begin{array}{rlrl} X\; : & 1:0:0 \hspace{2cm} & Y\;: & 0:1:0 \\ Z\;: & 0:0:1 & P\;: & \mathrm{cosec}\,(X + \tfrac12\pi):\mathbf{cosec}\,(Y + \tfrac13\pi):\mathrm{cosec}(Z + \tfrac13\pi) \; = \; \tfrac{14}{5\sqrt{3}}:\tfrac{2\sqrt{13}}{3\sqrt{3}}:\tfrac{14\sqrt{13}}{23\sqrt{3}} \end{array}$ Doing the standard calculations, the exact trilinear coordinates of these four points are (note that the area of the triangle is $\Delta = \tfrac12yz\sin X = 6\sqrt{3}$ ): $\begin{array}{rlrl} X\; : & \tfrac{6\sqrt{3}}{\sqrt{13}}:0:0 \hspace{2cm} & Y\;: & 0:\tfrac{12\sqrt{3}}{7}:0 \\ Z\;: & 0:0:4\sqrt{3} & P\;: & \tfrac{261\sqrt{3}}{91\sqrt{13}}:\tfrac{435\sqrt{3}}{637}:\tfrac{45\sqrt{3}}{91} \end{array}$ Since the distance between two points with (exact) trilinear coordinates $a_1:b_1:c_1$ and $a_2:b_2:c_2$ is given by the (amongst many others) the formula $\frac{\sqrt{(a_1-a_2)^2 + (b_1-b_2)^2 + 2(a_1-a_2)(b_1-b_2)\cos Z}}{\sin Z}$ we deduce the distances $a \; = \; PX \; = \; \tfrac{15}{\sqrt{91}} \hspace{1cm} b \ = \; PY \; = \; \tfrac{18}{\sqrt{91}} \hspace{1cm} c \; = \; PZ \; = \; \tfrac{58}{\sqrt{91}}$ and these values can be readily checked as satisfying the original conditions.

Note that $49 \times 18^2 - 33 \times 18 \times 58 + 9 \times 58^2 \; = \; 52 \times 15^2$ and hence $\boxed{\displaystyle \frac{49b^2 - 33bc + 9c^2}{a^2} \; = \; 52}$ It is worth noting that $25bc = 116a^2$ , and hence that $\frac{49b^2 + (25n - 33)bc + 9c^2}{a^2} \; = \; 52 + 116n$ for any integer $n$ . In particular, when $n=3$ , $\displaystyle\frac{49b^2 + 42bc + 9c^2}{a^2} \; = \; 400$ Since it is also true that $7 \times 18 + 3\times 58 \; = \; 20 \times 15$ we have $7b + 3c = 20a$ , and hence many other identities are possible.

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Here is another approach.

If we start with the equations $\begin{array}{rclll} a^2 + ab + b^2 & = & 9 & \hspace{3cm} & (1) \\ b^2 + bc + c^2 & = & 52 & & (2) \\ a^2 + ac + c^2 & = & 49 & & (3) \end{array}$ then subtracting these equations in pairs, we obtain $(c - b)(a + b + c) \; = \; 40 \hspace{2cm} (b - a)(a + b + c) \; = \; 3$ and hence $3(c-b) = 40(b-a)$ , so that $40a + 3c \; = \; 43b$ Thus we can substitute for $b$ in equation $(1)$ , obtaining $a^2 + a\left(\frac{40a + 3c}{43}\right) + \left(\frac{40a + 3c}{43}\right)^2 \; = \; 9$ which simplifies to $\begin{array}{rclll} 5169a^2 + 369ac + 9c^2 & = & 16641 & \hspace{3cm} & (4) \end{array}$ Thus $49(4) - 16641(3)$ gives $\begin{aligned} 120(1972a^2 + 12ac - 135c^2) & = \; 0 \\ 120(34a + 9c)(58a - 15c) & = \; 0 \end{aligned}$ If $c = \tfrac{58}{15}a$ then equation $(3)$ tells us that $a = \pm \tfrac{15}{\sqrt{91}}$ , and hence we obtain the solutions $a \; = \; \pm\frac{15}{\sqrt{91}} \hspace{1cm} b \; = \; \pm\frac{18}{\sqrt{91}} \hspace{1cm} c \; = \; \pm\frac{58}{\sqrt{91}}$ If, on the other hand, $c = -\tfrac{34}{9}a$ , then equation $(3)$ tells us that $a = \pm\tfrac{9}{\sqrt{19}}$ , and hence we obtain the solutions $a \; = \; \pm\frac{9}{\sqrt{19}} \hspace{1cm} b \; = \; \pm\frac{6}{\sqrt{19}} \hspace{1cm} c \; = \; \mp\frac{34}{\sqrt{19}}$ Thus there is only one solution where $a,b,c$ are all positive (and one where all three values are negative), but there is another pair of solutions where both positive and negative values occur.

Indeed, this second class of solutions has a geometrical interpretation as well. We have $a = \pm QX \hspace{1cm} b = \pm QY \hspace{1cm} c = \mp QZ$ where $Q$ is the Second Fermat point of the triangle (note that $\angle XQZ = \angle YQZ = 60^\circ$ as well as $\angle XQY = 120^\circ$ , which accounts for the sign changes).

Here was my solution. $b^2+bc+c^2-a^2-ab-b^2=52-9 \Rightarrow bc-ab+c^2-a^2=43\Rightarrow (c-a)(a+b+c)=43$ $b^2+bc+c^2-c^2-ca-a^2=52-49\Rightarrow bc-ca+b^2-a^2=3\Rightarrow (b-a)(a+b+c)=3$ $\Rightarrow 43(b-a)(a+b+c)=3(c-a)(a+b+c)\Rightarrow 43b-43a=3c-3a\Rightarrow c=\frac{43b-40a}{3}$ $\Rightarrow c^2+ac+a^2=\Big(\frac{43b-40a}{3}\Big)^2+a\Big(\frac{43b-40a}{3}\Big)+a^2=49$ $\Rightarrow \Big(\frac{43}{3}\Big)^2b^2+\Big(\frac{43}{3}-\frac{80\cdot 43}{9}\Big)ab+\Big(\Big(\frac{40}{3}\Big)^2-\frac{40}{3}+1\Big)=49$ $\Big(\frac{43}{3}\Big)^2b^2+\Big(\frac{43}{3}-\frac{80\cdot 43}{9}\Big)ab+\Big(\Big(\frac{40}{3}\Big)^2-\frac{40}{3}+1\Big)-\Big(\frac{43}{3}\Big)^2(a^2+ab+b^2)=49-9\Big(\frac{43}{3}\Big)^2$ $\Rightarrow \frac{1720}{3}ab+40a^2=1800$ $\Rightarrow b=\frac{1}{43}\Big(\frac{135}{a}-3a\Big)$ $\Rightarrow a^2+ab+b^2=a^2+a\Big[\frac{1}{43}\Big(\frac{135}{a}-3a\Big)\Big]+\Big[\frac{1}{43}\Big(\frac{135}{a}-3a\Big)\Big]^2=9$ $\Rightarrow 1729a^4-11646a^2+135^2=0 \Rightarrow a^2=\frac{225}{91},\frac{81}{19}$ Recall the problem asks for positive values of $a,b,c$ . There is only one positive value of $a$ that produces positive values for both $b$ and $c$ . Whence, the sought solution is $a=\frac{15}{\sqrt{91}},b=\frac{18}{\sqrt{91}},c=\frac{58}{\sqrt{91}}$ . I'll leave it for the reader to verify that this leads to the answer, 52.