A straight line \(L\) through the point \((3,-2)\) is inclined at an angle \(60^{\circ}\) to the line \(\sqrt{3} x+y=1\). If \(L\) also intersects the \(X\)-axis, then the equation of \(L\) is

A straight line passing through \(P\) and making an angle of \(\alpha=60^{\circ}\), is given by \(\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)\)


where, \(\quad \sqrt{3} x+y=1\)
\[
\Rightarrow \quad y=-\sqrt{3} x+1
\]
Then, \(\quad \tan \theta=-\sqrt{3}\)
\[
\begin{array}{ll}
\Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\
\Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\
\text { and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}
\end{array}
\]

\[
\begin{array}{ll}
\Rightarrow & y+2=0 \\
\text { and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\
\Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3}
\end{array}
\]
Neglecting, \(y+2=0\) as it does not intersect \(Y\)-axis.