Solution:

Place $A = (0, \sqrt{3}),$ $B = (-1, 0),$ $C = (1, 0),$ so $D = (0, 0),$ $E = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $F = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ The $120^\circ$ rotation about the center $G$ sends $A \to B \to C$ and $D \to E \to F,$ so we may take the symmetric configuration $P = D + t(E - D),$ $Q = E + t(F - E),$ $R = F + t(D - F);$ the rotation then carries the condition "$P$ on $\overline{CQ}$" to the other two conditions, so it suffices to make $C,$ $P,$ $Q$ collinear.

With $P = \left(\frac{t}{2}, \frac{t\sqrt{3}}{2}\right)$ and $Q = \left(\frac{1}{2} - t, \frac{\sqrt{3}}{2}\right),$ the cross product of $\overrightarrow{CP}$ and $\overrightarrow{CQ}$ is a multiple of $1 - t - t^2,$ so $t^2 + t - 1 = 0$ and $t = \frac{\sqrt{5} - 1}{2}.$ Both triangles are equilateral with center $G = \left(0, \frac{\sqrt{3}}{3}\right),$ so the area ratio is $\frac{GA^2}{GP^2}.$ Using $t^2 = 1 - t,$ $$GP^2 = \frac{t^2}{4} + 3\left(\frac{t}{2} - \frac{1}{3}\right)^2 = t^2 - t + \frac{1}{3} = \frac{7}{3} - \sqrt{5}, \qquad GA^2 = \frac{4}{3}.$$

Hence $$\frac{[ABC]}{[PQR]} = \frac{4/3}{7/3 - \sqrt{5}} = \frac{4}{7 - 3\sqrt{5}} = 7 + 3\sqrt{5},$$ so $a = 7,$ $b = 3,$ $c = 5,$ and $a^2 + b^2 + c^2 = 49 + 9 + 25 = 83.$