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(A) 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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(A) 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: $\frac{x-3}{\cos(\pi/6)} = \frac{y-4}{\sin(\pi/6)} = r$,where $r$ is the distance $PQ$.Thus,$x = 3 + r \cos(\pi/6) = 3 + r \frac{\sqrt{3}}{2}$ and $y = 4 + r \sin(\pi/6) = 4 + \frac{r}{2}$.Since $Q$ lies on the line $12x+5y+10=0$,we substitute these coordinates into the equation:$12(3 + \frac{r\sqrt{3}}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$$36 + 6r\sqrt{3} + 20 + 2.5r + 10 = 0$$66 + r(6\sqrt{3} + 2.5) = 0$$r(6\sqrt{3} + 2.5) = -66$Since $r$ represents a length,we take the absolute value:$r = \frac{66}{6\sqrt{3} + 2.5} = \frac{132}{12\sqrt{3} + 5}$.

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

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