If the straight line passes through the point P(3,4) makes an angle \[\dfrac{\pi }{6}\] with the x-axis and meets the line 12x + 5y +10 = 0 at Q, then length PQ is
(A) \[\dfrac{132}{12\sqrt{3}+5}\]
(B) \[\dfrac{132}{12\sqrt{3}-5}\]
(C) \[\dfrac{132}{12\sqrt{3}+12}\]
(D) \[\dfrac{132}{12\sqrt{3}-12}\]

Hint: To find the distance between PQ. First we have to write the equation of line passing through P in distance form. Then we have to substitute the point obtained from distance form in the given line equation.

Complete step-by-step answer:
Given that \[\theta =\dfrac{\pi }{6}\] be the inclination of the line with x-axis.
We know the formula for the equation of line passing through point \[\left( {{x}_{1}},{{y}_{1}} \right)\] in distance form is \[\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r\].
Substituting \[\left( {{x}_{1}},{{y}_{1}} \right)\] as (3,4) and \[\theta =\dfrac{\pi }{6}\], applying the above formula we get,
Equation of line passing through the point P(3,4) in distance form is \[\dfrac{x-3}{\cos \dfrac{\pi }{6}}=\dfrac{y-4}{\sin \dfrac{\pi }{6}}=r\].
Simplifying the expression we get,
\[x=r\cos \dfrac{\pi }{6}+3\],\[y=r\sin \dfrac{\pi }{6}+4\]

Therefore any point on the line is \[\left( 3+r\cos \dfrac{\pi }{6},4+rsin\dfrac{\pi }{6} \right)\]
Let this point be Q
Substituting the value of \[\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}\] and \[sin\dfrac{\pi }{6}=\dfrac{1}{2}\] in point Q we get,
\[Q=\left( 3+\dfrac{\sqrt{3}r}{2},4+\dfrac{r}{2} \right)\]
Solving point Q to get a simplified value we get,
\[Q=\left( \dfrac{\sqrt{3}r+6}{2},\dfrac{r+8}{2} \right)\]
Since Q lies on the line 12x+5y+10 = 0
Substituting the values of point Q in the equation we get,
\[\begin{align}
  & \dfrac{12\left( \sqrt{3}r+6 \right)}{2}+\dfrac{5\left( r+8 \right)}{2}+10=0 \\
 & 12\sqrt{3}r+72+5r+40+20=0 \\
 & \left( 12\sqrt{3}+5 \right)r+132=0
\end{align}\]
Subtracting with 132 from both sides we get,
\[(12\sqrt{3}+5)r=-132\]
Dividing with \[(12\sqrt{3}+5)\] on both sides we get,
\[r=\dfrac{-132}{(12\sqrt{3}+5)}\]
Since distance r cannot be negative we take it as a positive value.
Hence \[r=PQ=\dfrac{132}{(12\sqrt{3}+5)}\] where r is the distance between P and Q.

So, the correct answer is “Option A”.

Note: We will get a negative value of ‘r’, as we all know that the distance cannot be negative. Distance is always positive. We have to apply modulus then the distance becomes positive. Take care while doing calculations.
Alternative solution tip: If a straight line passes through P\[\left( {{x}_{1}},{{y}_{1}} \right)\] having an inclination \[\theta \] meets the line ax + by + c =0 at Q, then the distance between P and Q is given by
\[PQ=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\left| a\cos \theta +b\sin \theta \right|}\]