Let $$A(\alpha, 0)$$ and $$B(0, \beta)$$ be the points on the line $$5x + 7y = 50$$. Let the point $$P$$ divide the line segment $$AB$$ internally in the ratio $$7:3$$. Let $$3x - 25 = 0$$ be a directrix of the ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ and the corresponding focus be $$S$$. If from $$S$$, the perpendicular on the $$x$$-axis passes through $$P$$, then the length of the latus rectum of $$E$$ is equal to

Line $$5x + 7y = 50$$: $$A(\alpha,0) \Rightarrow 5\alpha = 50 \Rightarrow \alpha = 10$$. $$B(0,\beta) \Rightarrow 7\beta = 50 \Rightarrow \beta = 50/7$$.

P divides AB in ratio 7:3 internally: $$P = \left(\frac{7(0)+3(10)}{10}, \frac{7(50/7)+3(0)}{10}\right) = (3, 5)$$.

Directrix: $$3x - 25 = 0 \Rightarrow x = 25/3$$. For ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$: directrix $$x = a/e$$.

So $$a/e = 25/3$$. Focus $$S = (ae, 0)$$.

Since the perpendicular from S on x-axis passes through P, the x-coordinate of S = x-coordinate of P = 3. So $$ae = 3$$.

From $$a/e = 25/3$$ and $$ae = 3$$: $$a^2 = (a/e)(ae) = (25/3)(3) = 25$$, so $$a = 5$$.

$$e = ae/a = 3/5$$. $$b^2 = a^2(1-e^2) = 25(1-9/25) = 16$$.

Length of latus rectum = $$\frac{2b^2}{a} = \frac{32}{5}$$.

The answer is Option (4): $$\boxed{\frac{32}{5}}$$.