Let $$R$$ be the point $$(3, 7)$$ and let $$P$$ and $$Q$$ be two points on the line $$x + y = 5$$ such that $$PQR$$ is an equilateral triangle. Then the area of $$\triangle PQR$$ is

We need to find the area of equilateral triangle $$PQR$$ where $$R = (3, 7)$$ and $$P, Q$$ lie on the line $$x + y = 5$$.

Since the height of the equilateral triangle from vertex $$R(3, 7)$$ to the base $$PQ$$ is given by the point-to-line distance formula $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ for the line $$x + y - 5 = 0$$, substituting yields $$d = \frac{|1(3) + 1(7) - 5|}{\sqrt{1^2 + 1^2}} = \frac{|3 + 7 - 5|}{\sqrt{2}} = \frac{5}{\sqrt{2}}$$. This distance $$d$$ is thus the height of the triangle from vertex $$R$$ to the base $$PQ$$.

Now, for an equilateral triangle with side length $$s$$ the height satisfies $$h = \frac{s\sqrt{3}}{2}$$, and setting $$h = d$$ gives $$\frac{s\sqrt{3}}{2} = \frac{5}{\sqrt{2}}$$. Solving for $$s$$ gives $$s = \frac{10}{\sqrt{2} \cdot \sqrt{3}} = \frac{10}{\sqrt{6}}$$.

From this, because $$PQ$$ lies along the line $$x + y = 5$$ and has length $$s = \frac{10}{\sqrt{6}}$$, and the foot of the perpendicular from $$R$$ to the line is $$\left(\frac{1}{2}, \frac{9}{2}\right)$$, the points $$P$$ and $$Q$$ are situated symmetrically at distance $$\frac{s}{2}$$ from this foot along the line, confirming that both points lie on the line and the triangle is valid.

Next, the area of the equilateral triangle is given by $$\text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times \frac{100}{6} = \frac{100\sqrt{3}}{24} = \frac{25\sqrt{3}}{6}$$.

Therefore, rationalizing shows that $$\frac{25\sqrt{3}}{6} = \frac{25}{2\sqrt{3}}$$, so the area equals $$\frac{25}{2\sqrt{3}}$$.

The correct answer is Option D.