$\begin{array}{l} I f t h e l i n e y = \sqrt{3} x c u t s t h e c u r v e x^{3} + y^{3} + 3 x y + 5 x^{2} + 3 y^{2} + 4 x + 5 y - 1 = 0 a t t h e p o int s A, B, C t h e n O A . O B . O C i s \end{array}$

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$\frac{4}{13} (3 \sqrt{3} - 1)$

$(3 \sqrt{3} - 1)$

$\frac{2}{\sqrt{3}} + 7$

$\frac{14}{13} (3 \sqrt{3} + 1)$

answer is A.

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Detailed Solution

$\begin{array}{l} L e t t h e p o int o n t h e l i n e y = \sqrt{3} x i s \\ (r \cos θ, r \sin θ) a n d m = \sqrt{3} \\ T a n θ = \sqrt{3} \Rightarrow θ = 60^{0} \\ ∴ T h e p o int (r \cos 60^{0}, r \sin 60^{0}) \\ = (\frac{r}{2}, \frac{\sqrt{3} r}{2}) \\ x^{3} + y^{3} + 3 x y + 5 x^{2} + 3 y^{2} + 4 x + 5 y - 1 = 0 \\ \Rightarrow \frac{r^{3}}{8} + \frac{3 \sqrt{3} r^{3}}{8} + \frac{3 \sqrt{3} r^{2}}{4} + 5 (\frac{r^{2}}{4}) + 3 (\frac{3 r^{2}}{4}) + 4 (\frac{r}{2}) + 5 (\frac{\sqrt{3} r}{2}) - 1 = 0 \\ \Rightarrow (\frac{1 + 3 \sqrt{3}}{8}) r^{3} + \frac{r^{2}}{4} (3 \sqrt{3} + 14) + \frac{r}{2} (5 \sqrt{3} + 4) - 1 = 0 \end{array}$

$\begin{array}{l} ∴ r_{1} r_{2} r_{3} = \frac{- (- 1)}{(\frac{1 + 3 \sqrt{3}}{8})} = \frac{8}{1 + 3 \sqrt{3}} \\ = \frac{8}{3 \sqrt{3} + 1} \times \frac{3 \sqrt{3} - 1}{3 \sqrt{3} - 1} \\ = \frac{8 (3 \sqrt{3} - 1)}{27 - 1} \\ = \frac{8 (3 \sqrt{3} - 1)}{26} \\ = \frac{4 (3 \sqrt{3} - 1)}{13} \end{array}$