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(C) The equation of the line passing through $P(3,4)$ with an angle of inclination $	heta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x-

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$\frac{132}{12\sqrt{3} + 5}$

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(C) The equation of the line passing through $P(3,4)$ with an angle of inclination $	heta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x-3}{\cos(\pi/6)} = \frac{y-4}{\sin(\pi/6)} = r$. Substituting $\cos(\pi/6) = \frac{\sqrt{3}}{2}$ and $\sin(\pi/6) = \frac{1}{2}$,we get: $\frac{x-3}{\sqrt{3}/2} = \frac{y-4}{1/2} = r$. Thus,any point $Q$ on this line is given by $(3 + \frac{\sqrt{3}r}{2}, 4 + \frac{r}{2})$. Since $Q$ lies on the line $12x + 5y + 10 = 0$,we substitute the coordinates of $Q$ into this equation: $12(3 + \frac{\sqrt{3}r}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$. $36 + 6\sqrt{3}r + 20 + 2.5r + 10 = 0$. $66 + (6\sqrt{3} + 2.5)r = 0$. $66 + (\frac{12\sqrt{3} + 5}{2})r = 0$. $r = -\frac{132}{12\sqrt{3} + 5}$. The length $PQ$ is $|r| = \frac{132}{12\sqrt{3} + 5}$.

If the straight line passing through the point $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the positive direction of $X$-axis and meets the line $12x + 5y + 10 = 0$ at $Q$,then the length of $PQ$ is

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