A variable line $$L$$ passes through the point $$(3, 5)$$ and intersects the positive coordinate axes at the points A and B. The minimum area of the triangle OAB, where O is the origin, is :

A line through $$(3, 5)$$ with intercepts on positive axes at A$$(a, 0)$$ and B$$(0, b)$$.

Line: $$\frac{x}{a} + \frac{y}{b} = 1$$. Since it passes through $$(3, 5)$$: $$\frac{3}{a} + \frac{5}{b} = 1$$.

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

We need to minimize $$ab$$ subject to $$\frac{3}{a} + \frac{5}{b} = 1$$ with $$a > 3, b > 5$$ (positive intercepts with point (3,5) between them).

By AM-GM: $$1 = \frac{3}{a} + \frac{5}{b} \geq 2\sqrt{\frac{15}{ab}}$$.

$$\frac{1}{4} \geq \frac{15}{ab} \implies ab \geq 60$$.

Minimum area = $$\frac{1}{2} \times 60 = 30$$.

Equality when $$\frac{3}{a} = \frac{5}{b}$$, i.e., $$a = 6, b = 10$$.

The correct answer is Option 1: 30.