Let R be the point 3 , 7 and let P and Q be two points on the line x + y = 5 such that P Q R is an…

Given $PQR$ is an equilateral triangle.

Let the side of the triangle be $a$

Perpendicular distance from $R\left(3,7\right)$ to the line $x+y=5$ is

$\frac{\left|3+7-5\right|}{\sqrt{1+1}}=\frac{5}{\sqrt{2}}$

So $\sin 60°=\frac{{\displaystyle \frac{5}{\sqrt{2}}}}{a}$

$a=\frac{5\sqrt{2}}{\sqrt{3}}$

Area of $∆PQR=\frac{\sqrt{3}}{4}{a}^2=\frac{25}{2\sqrt{3}}$