The joint equation of pair of lines through the origin and making an angle of \(\frac{\pi}{6}\) with the line \(3 x+y-6=0\) is

The slope of the line \(3 x+y-6=0\) is \(m_1=-3\). Let \(m\) be the slope of one of the lines making an angle \(\frac{\pi}{6}\) with \(3 x+y-6=0\).
\(\begin{aligned}
& \therefore \quad \tan \frac{\pi}{6}=\left|\frac{\mathrm{m}-\mathrm{m}_1}{1+\mathrm{m}_1}\right| \\
& \quad \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{\mathrm{m}-(-3)}{1+\mathrm{m}(-3)}\right| \\
& \quad \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{\mathrm{m}+3}{1-3 \mathrm{~m}}\right|
\end{aligned}\)
Squaring on both sides, we get
\(\begin{aligned}
& (1-3 m)^2=3(m+3)^2 \\
& \Rightarrow 6 m^2-24 m-26=0 \\
& \Rightarrow 3 m^2-12 m-13=0
\end{aligned}\)
This is the auxiliary equation of two lines and their joint equation is obtained by putting \(\mathrm{m}=\frac{y}{x}\).
\(\therefore \quad\) The joint equation of the lines is
\(\begin{aligned}
& 3\left(\frac{y}{x}\right)^2-12\left(\frac{y}{x}\right)-13=0 \\
& \Rightarrow 13 x^2+12 x y-3 y^2=0
\end{aligned}\)