Let a triangle be bounded by the lines $$L_1: 2x + 5y = 10$$; $$L_2: -4x + 3y = 12$$ and the line $$L_3$$, which passes through the point $$P(2,3)$$, intersect $$L_2$$ at $$A$$ and $$L_1$$ at $$B$$. If the point $$P$$ divides the line-segment $$AB$$, internally in the ratio 1 : 3, then the area of the triangle is equal to

Given lines: $$L_1: 2x + 5y = 10$$, $$L_2: -4x + 3y = 12$$, and $$L_3$$ passes through $$P(2, 3)$$.

Point $$A$$ is on $$L_2$$, point $$B$$ is on $$L_1$$, and $$P$$ divides $$AB$$ internally in the ratio $$1:3$$.

Find the coordinates of A and B:

Let $$A = \left(a,\; \frac{12 + 4a}{3}\right)$$ on $$L_2$$ and $$B = \left(b,\; \frac{10 - 2b}{5}\right)$$ on $$L_1$$.

Using the section formula (P divides AB in ratio 1:3):

$$2 = \frac{3a + b}{4} \implies 3a + b = 8 \quad \cdots (1)$$

$$3 = \frac{3 \cdot \frac{12 + 4a}{3} + \frac{10 - 2b}{5}}{4} = \frac{(12 + 4a) + \frac{10 - 2b}{5}}{4}$$

$$12 = 12 + 4a + \frac{10 - 2b}{5} = 12 + 4a + 2 - \frac{2b}{5}$$

$$0 = 4a + 2 - \frac{2b}{5}$$

$$20a + 10 - 2b = 0 \implies 10a - b = -5 \quad \cdots (2)$$

Adding (1) and (2): $$13a = 3 \implies a = \frac{3}{13}$$

From (1): $$b = 8 - \frac{9}{13} = \frac{95}{13}$$

$$A = \left(\frac{3}{13},\; \frac{12 + \frac{12}{13}}{3}\right) = \left(\frac{3}{13},\; \frac{56}{13}\right)$$

$$B = \left(\frac{95}{13},\; \frac{10 - \frac{190}{13}}{5}\right) = \left(\frac{95}{13},\; -\frac{12}{13}\right)$$

Find the intersection of $$L_1$$ and $$L_2$$:

Solving $$2x + 5y = 10$$ and $$-4x + 3y = 12$$ simultaneously:

Multiply the first equation by 2: $$4x + 10y = 20$$

Adding: $$13y = 32 \implies y = \frac{32}{13}$$

$$x = \frac{10 - \frac{160}{13}}{2} = \frac{-\frac{30}{13}}{2} = -\frac{15}{13}$$

So $$C = \left(-\frac{15}{13},\; \frac{32}{13}\right)$$

Calculate the area of triangle ABC:

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

$$= \frac{1}{2}\left|\frac{3}{13}\left(-\frac{12}{13} - \frac{32}{13}\right) + \frac{95}{13}\left(\frac{32}{13} - \frac{56}{13}\right) + \left(-\frac{15}{13}\right)\left(\frac{56}{13} + \frac{12}{13}\right)\right|$$

$$= \frac{1}{2}\left|\frac{3}{13} \cdot \frac{-44}{13} + \frac{95}{13} \cdot \frac{-24}{13} + \frac{-15}{13} \cdot \frac{68}{13}\right|$$

$$= \frac{1}{2}\left|\frac{-132 - 2280 - 1020}{169}\right| = \frac{1}{2} \cdot \frac{3432}{169} = \frac{1716}{169} = \frac{132}{13}$$

The correct answer is Option B: $$\dfrac{132}{13}$$.