Answer: 120

Small Hint:

Since $15^2 + 20^2 = 25^2,$ the triangle has a right angle at $R$

Big Hint:

Subtract the three corner triangles from $\triangle PQR;$ each corner triangle has two sides of length $5$ around a known angle, so its area is $\frac{1}{2} \cdot 25 \cdot \sin(\text{angle})$

Solution:

Since $15^2 + 20^2 = 25^2,$ the triangle is right-angled at $R,$ and its area is $\frac{1}{2} \cdot 15 \cdot 20 = 150.$ Also $\sin P = \frac{20}{25} = \frac{4}{5}$ and $\sin Q = \frac{15}{25} = \frac{3}{5}.$

The hexagon is the triangle minus three corner triangles, each with two sides of length $5:$ at $P,$ area $\frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{4}{5} = 10;$ at $Q,$ area $\frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{3}{5} = \frac{15}{2};$ at $R,$ area $\frac{1}{2} \cdot 5 \cdot 5 = \frac{25}{2}.$

Therefore the hexagon has area $150 - 10 - \frac{15}{2} - \frac{25}{2} = 120.$