A point $P$ lies on the image parabola exactly when its rotation by $-60^\circ,$ namely $$Q = \left(\frac{x + \sqrt{3}y}{2},\ \frac{-\sqrt{3}x + y}{2}\right),$$ lies on the original parabola. So we need $P$ and $Q$ both on $y = x^2 - 4.$ The parabola is symmetric in $x \mapsto -x,$ so we look for $P = (x, y)$ whose rotated image is the mirror point $Q = (-x, y).$

Matching $y$-coordinates gives $-\frac{\sqrt{3}}{2}x + \frac{y}{2} = y,$ i.e. $y = -\sqrt{3}\,x,$ and then the $x$-coordinate works automatically: $\frac{x}{2} + \frac{\sqrt{3}}{2}(-\sqrt{3}x) = -x.$ Substituting $y = -\sqrt{3}\,x$ into $y = x^2 - 4$ gives $x^2 + \sqrt{3}\,x - 4 = 0,$ whose positive root is $x = \frac{-\sqrt{3} + \sqrt{19}}{2}.$ Then $$y = -\sqrt{3}\,x = \frac{3 - \sqrt{57}}{2} \lt 0,$$ so this point is in the fourth quadrant, on both curves.

The problem guarantees the fourth-quadrant intersection is unique, so its $y$-coordinate is $\frac{3 - \sqrt{57}}{2},$ giving $a + b + c = 3 + 57 + 2 = 62.$