If a straight line through the point P(3, 5) makes an angle $\frac{π}{6}$ with +ve x-axis and meets the lines $2 x + y + 5 = 0$ $& 3 x - 2 y + 7 = 0$ at Q & R then $\frac{P Q}{P R} =$

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$\frac{3 \sqrt{3} - 2}{2 \sqrt{3} + 1}$

$\frac{(6 \sqrt{3} + 3)}{(12 \sqrt{3} - 8)}$

$\frac{2 (12 \sqrt{3} - 8)}{6 \sqrt{3} + 3}$

$\frac{3 \sqrt{3} + 1}{3 \sqrt{3} - 2}$

answer is C.

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Detailed Solution

Here line is $\frac{x - 3}{\sqrt{3} / 2} = \frac{y - 5}{1 / 2} = r (let)$

$∴ Q (3 + \frac{\sqrt{3} r_{1}}{2}, 5 + \frac{r_{1}}{2}), R (3 + \frac{r_{1} \sqrt{3}}{2}, 5 + \frac{r_{2}}{2})$

$∴ 2 (3 + \frac{\sqrt{3} r_{1}}{2}) + 5 + \frac{r_{1}}{2} + 5 = 0$ and $3 (3 + \frac{r_{2} \sqrt{3}}{2}) - 2 (5 + \frac{r_{1}}{2}) + 7 = 0$

$\Rightarrow r_{1} = \frac{- 32}{2 \sqrt{3} + 1} & r_{2} = \frac{- 12}{3 \sqrt{3} - 2}$

$∴ \frac{P Q}{P R} = \frac{24 \sqrt{3} - 16}{6 \sqrt{3} + 3}$

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