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(C) The slope of the line $3x+y-6=0$ is $m_1 = -3$. Let $m$ be the slope of the lines making an angle $	heta = \frac{\pi}{6}$ with the given line. Us

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$13x^2+12xy-3y^2=0$

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(C) The slope of the line $3x+y-6=0$ is $m_1 = -3$. Let $m$ be the slope of the lines making an angle $	heta = \frac{\pi}{6}$ with the given line. Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$,we have: $\frac{1}{\sqrt{3}} = \left| \frac{m - (-3)}{1 + m(-3)} ight| = \left| \frac{m+3}{1-3m} ight|$. Squaring both sides: $\frac{1}{3} = \frac{(m+3)^2}{(1-3m)^2} \Rightarrow (1-3m)^2 = 3(m+3)^2$. $1 - 6m + 9m^2 = 3(m^2 + 6m + 9) = 3m^2 + 18m + 27$. $6m^2 - 24m - 26 = 0 \Rightarrow 3m^2 - 12m - 13 = 0$. Substituting $m = \frac{y}{x}$ to find the joint equation: $3(\frac{y}{x})^2 - 12(\frac{y}{x}) - 13 = 0$. Multiplying by $x^2$: $3y^2 - 12xy - 13x^2 = 0 \Rightarrow 13x^2 + 12xy - 3y^2 = 0$.

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

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