1. Given data

• $R=(3,7)$
• $P,Q$ lie on the line $x+y=5$
• $\triangle PQR$ is equilateral

Since $P$ and $Q$ are on the line $x+y=5$, the side $PQ$ lies on this line.

2. Use the altitude property of an equilateral triangle

In an equilateral triangle of side $a$, the altitude is

$h=\frac{\sqrt{3}}{2}a$

So if we know the perpendicular distance from $R$ to the line containing $PQ$, we can find the side length.

3. Find the perpendicular distance from $R(3,7)$ to the line $x+y-5=0$

Distance from $(x_1,y_1)$ to $Ax+By+C=0$ is

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

Here, $A=1$, $B=1$, $C=-5$, and $(x_1,y_1)=(3,7)$.

Thus,

$d=\frac{|3+7-5|}{\sqrt{1^2+1^2}}=\frac{5}{\sqrt{2}}$

This is the altitude of the equilateral triangle:

$h=\frac{5}{\sqrt{2}}$

4. Find the side length $a$

Using

$h=\frac{\sqrt{3}}{2}a$

we get

$a=\frac{2h}{\sqrt{3}}=\frac{2}{\sqrt{3}}\cdot \frac{5}{\sqrt{2}}=\frac{10}{\sqrt{6}}$

5. Find the area

Area of an equilateral triangle is

$\text{Area}=\frac{\sqrt{3}}{4}a^2$

Now,

$a^2=\left(\frac{10}{\sqrt{6}}\right)^2=\frac{100}{6}=\frac{50}{3}$

So,

$\text{Area}=\frac{\sqrt{3}}{4}\cdot \frac{50}{3}=\frac{50\sqrt{3}}{12}=\frac{25\sqrt{3}}{6}$

Rationalizing the denominator form given in options,

$\frac{25\sqrt{3}}{6}=\frac{25}{2\sqrt{3}}$

because

$\frac{25}{2\sqrt{3}}=\frac{25\sqrt{3}}{2\cdot 3}=\frac{25\sqrt{3}}{6}$

6. Match with options

This is option D.

$\boxed{\frac{25}{2\sqrt{3}}}$