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

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

We set up convenient coordinates for the vertices by letting $$P = (0, 0)$$, $$Q = (1, 0)$$, and $$R = (0, 1)$$.

$$\text{Area}(\triangle PQR) = \frac{1}{2}$$

Next, we find the coordinates of points $$A$$, $$B$$, and $$C$$ using the given ratios.

Since $$\frac{QA}{AR} = \frac{1}{2}$$, point $$A$$ divides segment $$QR$$ in the ratio $$1:2$$ from $$Q$$:

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

Since $$\frac{RB}{BP} = \frac{1}{2}$$, point $$B$$ divides segment $$RP$$ in the ratio $$1:2$$ from $$R$$:

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

Since $$\frac{PC}{CQ} = \frac{1}{2}$$, point $$C$$ divides segment $$PQ$$ in the ratio $$1:2$$ from $$P$$:

$$C = \frac{2P + 1 \cdot Q}{3} = \frac{(0,0) + (1,0)}{3} = \left(\frac{1}{3}, 0\right)$$

Then we calculate the area of triangle $$ABC$$ using the coordinate formula:

$$\text{Area} = \frac{1}{2}\bigl|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\bigr|$$

Substituting the coordinates:

$$= \frac{1}{2}\left|\frac{2}{3}\left(\frac{2}{3} - 0\right) + 0\left(0 - \frac{1}{3}\right) + \frac{1}{3}\left(\frac{1}{3} - \frac{2}{3}\right)\right|$$

$$= \frac{1}{2}\left|\frac{4}{9} + 0 - \frac{1}{9}\right| = \frac{1}{2} \times \frac{3}{9} = \frac{1}{6}$$

Finally, we compute the desired ratio:

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

Therefore, the answer is $$3$$.