Let a point A lie between the parallel lines $$L_{1}\text{ and }L_{2}$$ such that its distances from $$L_{1}\text{ and }L_{2}$$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines $$L_{1}\text{ and }L_{2}$$, respectively, is:

Point A lies between parallel lines L₁ and L₂ at distances 6 and 3 units from them respectively, and we seek the area of an equilateral triangle ABC with B on L₁ and C on L₂.

Since the total distance between L₁ and L₂ is 6 + 3 = 9 units, we use a coordinate approach: place L₁ at y = 6, L₂ at y = -3, and A at the origin. Then let B = (b, 6) and C = (c, -3).

We have $$AB^2 = b^2 + 36,\quad AC^2 = c^2 + 9,\quad BC^2 = (b-c)^2 + 81.$$ Equating AB = AC gives $$b^2 + 36 = c^2 + 9 \;\Rightarrow\; b^2 - c^2 = -27 \quad\text{(i)},$$ and equating AB = BC gives $$b^2 + 36 = (b-c)^2 + 81 \;\Rightarrow\; 2bc = c^2 + 45 \quad\text{(ii)}.$$

From (i): $$b^2 = c^2 - 27.$$ From (ii): $$b = \frac{c^2 + 45}{2c}.$$ Substituting into $$\left(\frac{c^2+45}{2c}\right)^2 = c^2 - 27$$ yields $$(c^2+45)^2 = 4c^2(c^2 - 27),$$ which expands to $$c^4 + 90c^2 + 2025 = 4c^4 - 108c^2,$$ and hence $$3c^4 - 198c^2 - 2025 = 0.$$

Solving gives $$c^2 = \frac{198 \pm \sqrt{198^2 + 4 \times 3 \times 2025}}{6} = \frac{198 \pm 252}{6},$$ and taking the positive root yields $$c^2 = 75.$$ Since $$AB^2 = c^2 - 27 + 36 = 75 + 9 = 84,$$ the area of the triangle is $$\frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}.$$

The answer is Option 3: $$21\sqrt{3}.$$