Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then $$(QR)^{2}$$ is equal to______.

Two parallel lines are 5 units apart and point P lies between them at 1 unit from one line. An equilateral triangle PQR is formed with Q on one line and R on the other. Find $$(QR)^2$$.

Choose coordinates so that the two parallel lines are $$y = 0$$ and $$y = 5$$, and take $$P = (0, 1)$$. Without loss of generality let $$Q = (a, 0)$$ on $$y = 0$$ and $$R = (b, 5)$$ on $$y = 5$$.

Equilateral triangle conditions give $$PQ = PR = QR$$, and hence $$PQ^2 = a^2 + 1,$$ $$PR^2 = b^2 + 16,$$ $$QR^2 = (b-a)^2 + 25.$$

Equating $$PQ^2 = PR^2$$ yields $$a^2 + 1 = b^2 + 16$$ so $$a^2 - b^2 = 15$$. Equating $$PQ^2 = QR^2$$ gives $$a^2 + 1 = (b-a)^2 + 25 = b^2 - 2ab + a^2 + 25$$, leading to $$1 = b^2 - 2ab + 25$$ and hence $$2ab = b^2 + 24$$.

From $$a^2 - b^2 = 15$$ we have $$a^2 = b^2 + 15$$, and from $$2ab = b^2 + 24$$ we get $$a = \frac{b^2 + 24}{2b}$$. Substituting into the expression for $$a^2$$ gives $$\left(\frac{b^2 + 24}{2b}\right)^2 = b^2 + 15,$$ so $$(b^2 + 24)^2 = 4b^2(b^2 + 15),$$ $$b^4 + 48b^2 + 576 = 4b^4 + 60b^2,$$ $$3b^4 + 12b^2 - 576 = 0,$$ $$b^4 + 4b^2 - 192 = 0.$$ Letting $$u = b^2$$ and applying the quadratic formula gives $$u = \frac{-4 + \sqrt{16 + 768}}{2} = \frac{-4 + 28}{2} = 12,$$ so $$b^2 = 12$$.

Since $$a^2 = b^2 + 15 = 27$$ we have $$a = 3\sqrt{3}$$ and $$b = 2\sqrt{3}$$, and one checks $$2ab = 2 \cdot 3\sqrt{3} \cdot 2\sqrt{3} = 36 = b^2 + 24 = 12 + 24 = 36$$. It follows that $$QR^2 = a^2 + 1 = 27 + 1 = 28$$.

The answer is 28.