Answer: 40

Solution:

The midpoint of $\overline{AB}$ is $M = \left(\frac{3}{2}, \sqrt{3}\right),$ and $AB = \sqrt{1 + 12} = \sqrt{13}.$ The third vertex lies at distance $\frac{\sqrt{3}}{2}\sqrt{13}$ from $M$ along a direction perpendicular to $\overrightarrow{AB} = (1, 2\sqrt{3});$ a unit perpendicular is $\frac{1}{\sqrt{13}}(2\sqrt{3}, -1).$ Taking the sign that lands in the first quadrant, $$\begin{aligned} C &= M + \frac{\sqrt{3}}{2}\,(2\sqrt{3}, -1) \\ &= \left(\frac{9}{2}, \frac{\sqrt{3}}{2}\right) \end{aligned}$$ (the other choice has negative $x$-coordinate).

The center of an equilateral triangle is its centroid, the average of the vertices: $$\begin{aligned} P &= \small \left(\frac{1 + 2 + \frac{9}{2}}{3},\ \frac{0 + 2\sqrt{3} + \frac{\sqrt{3}}{2}}{3} \right) \\ &= \left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right). \end{aligned}$$

Then $x \cdot y = \frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12},$ so $p + q + r = 25 + 3 + 12 = 40.$