Given PQ = 6 cm, QR = 8 cm, and the perimeter of PQR = 24 cm.

The perimeter of $\triangle PQR$ is the sum of its sides $PQ$, $QR$, and $PR$.
Given $PQ = 6 \text{ cm}$, $QR = 8 \text{ cm}$, and the perimeter of $\triangle PQR = 24 \text{ cm}$.

$$PQ + QR + PR = \text{Perimeter of } \triangle PQR$$ $$6 \text{ cm} + 8 \text{ cm} + PR = 24 \text{ cm}$$ $$14 \text{ cm} + PR = 24 \text{ cm}$$

Subtract $14 \text{ cm}$ from both sides:

$$PR = 24 \text{ cm} - 14 \text{ cm}$$ $$PR = 10 \text{ cm}$$

The area of a triangle is given by the formula:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

For $\triangle PRS$, we can consider $PR$ as the base. From the diagram, the perpendicular height from vertex $S$ to the base $PR$ is given as $3 \text{ cm}$.
Base ($PR$) $= 10 \text{ cm}$
Height $= 3 \text{ cm}$

$$\text{Area of } \triangle PRS = \frac{1}{2} \times PR \times \text{height}$$ $$\text{Area of } \triangle PRS = \frac{1}{2} \times 10 \text{ cm} \times 3 \text{ cm}$$ $$\text{Area of } \triangle PRS = 5 \text{ cm} \times 3 \text{ cm}$$ $$\text{Area of } \triangle PRS = 15 \text{ cm}^2$$

The area of $\triangle PRS$ is $\boxed{\text{15 cm}^2}$.