$$\def\o{\overline}\frac{\o{PS}}{\o{QS}} = \frac 3 1 \implies \frac{\o{PS} + \o{QS}}{\o{QS}} = \frac{3+1}{1} = 4 \implies \o{PQ} = 4\cdot \o{QS}$$
Similarly, $$\frac{\o{RT}}{\o{QT}} = \frac 5 2 \implies \frac{\o{RT} + \o{QT}}{\o{QT}} = \frac{5+2}{2} = \frac 7 2 \implies \o{RQ} = \frac 7 2 \cdot \o{QT}$$
Now, $$\text{Area of } \triangle PQR = \frac 1 2 \cdot \o{PQ} \cdot \o{RQ} = \frac 1 2 \cdot \left ( 4 \cdot \o{QS} \right ) \cdot \left ( \frac 7 2 \cdot {QT} \right ) \\= \frac 1 2 \cdot \left ( \o{QS} \cdot \o{QT} \right ) \cdot 4 \cdot \frac 7 2 = 20 \text{ cm}^2 \cdot 14 = 280 \text{ cm}^2$$