Let A ( α, 0 ) and B ( 0, β ) be the points on the line 5x + 7y =...

Given: $A ( \alpha, 0 )$ and $B ( 0, \beta )$ lies on the line $5x + 7y = 50$.

$⇒ 5\left(\alpha\right) + 7\left(0\right) = 50, 5\left(0\right) + 7\left(\beta\right) = 50$

$⇒ \alpha = 10, \beta = \dfrac{50}{7}$

Also, $P$ divide the line segment $AB$ internally in the ratio $7: 3$.

$⇒ P ≡ \left(\dfrac{7 \times 0 + 3 \times 10}{7 + 3}, \dfrac{7 \times \dfrac{50}{7} + 3 \times 0}{7 + 3}\right)$

$⇒ P ≡ \left(3, 5\right)$

$⇒ ae = 3$ as perpendicular from $S$ passes through $P$

Now, $3x - 25 = 0$ is a directrix of the ellipse $E: \dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

$⇒ x = \dfrac{25}{3}$

We know that, directrix of an ellipse is given by $x = \pm \dfrac{a}{e}$

$⇒ \dfrac{a}{e} = \dfrac{25}{3}$

Also, $ae = 3$

$⇒ a = 5, b = 4$

Length of LR $= \dfrac{2{b}^{2}}{a} = \dfrac{32}{5}$