Let $$PQR$$ be a triangle. The points $$A$$, $$B$$ and $$C$$ are on the sides $$QR$$, $$RP$$ and $$PQ$$ respectively such that $$\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$$. Then $$\frac{\text{Area}\triangle PQR}{\text{Area}\triangle ABC}$$ is equal to

Given triangle $$PQR$$ with points $$A$$, $$B$$, $$C$$ on sides $$QR$$, $$RP$$, $$PQ$$ respectively such that $$\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$$.

Let $$P = (0, 0)$$, $$Q = (1, 0)$$, $$R = (0, 1)$$.

Area of $$\triangle PQR = \frac{1}{2}$$.

Point $$A$$ on $$QR$$ with $$\frac{QA}{AR} = \frac{1}{2}$$:

$$A$$ divides $$QR$$ in ratio $$1:2$$ from $$Q$$.

$$A = Q + \frac{1}{3}(R - Q) = (1, 0) + \frac{1}{3}(-1, 1) = \left(\frac{2}{3}, \frac{1}{3}\right)$$

Point $$B$$ on $$RP$$ with $$\frac{RB}{BP} = \frac{1}{2}$$:

$$B$$ divides $$RP$$ in ratio $$1:2$$ from $$R$$.

$$B = R + \frac{1}{3}(P - R) = (0, 1) + \frac{1}{3}(0, -1) = \left(0, \frac{2}{3}\right)$$

Point $$C$$ on $$PQ$$ with $$\frac{PC}{CQ} = \frac{1}{2}$$:

$$C$$ divides $$PQ$$ in ratio $$1:2$$ from $$P$$.

$$C = P + \frac{1}{3}(Q - P) = (0, 0) + \frac{1}{3}(1, 0) = \left(\frac{1}{3}, 0\right)$$

Using the coordinate formula:

$$\text{Area}(\triangle ABC) = \frac{1}{2}\left|\det\begin{pmatrix} \frac{2}{3} - \frac{1}{3} & \frac{1}{3} - 0 \\ 0 - \frac{1}{3} & \frac{2}{3} - 0 \end{pmatrix}\right|$$

$$= \frac{1}{2}\left|\det\begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} \end{pmatrix}\right|$$

$$= \frac{1}{2}\left|\frac{1}{3} \cdot \frac{2}{3} - \frac{1}{3} \cdot \left(-\frac{1}{3}\right)\right|$$

$$= \frac{1}{2}\left|\frac{2}{9} + \frac{1}{9}\right| = \frac{1}{2} \cdot \frac{3}{9} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$$

$$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \frac{1/2}{1/6} = 3$$

Therefore, the answer is Option B: $$\mathbf{3}$$.