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(D) Let the lines passing through the origin be $y = mx$. The slope of the line $3x + y - 6 = 0$ is $m_1 = -3$.The angle $	heta = \frac{\pi}{4}$ betw

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

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(D) Let the lines passing through the origin be $y = mx$. The slope of the line $3x + y - 6 = 0$ is $m_1 = -3$.The angle $	heta = \frac{\pi}{4}$ between the lines $y = mx$ and $y = -3x + 6$ is given by $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$.Substituting the values: $	an \frac{\pi}{4} = \left| \frac{m - (-3)}{1 + m(-3)} ight| \implies 1 = \left| \frac{m + 3}{1 - 3m} ight|$.This gives two cases: $1 = \frac{m + 3}{1 - 3m}$ or $-1 = \frac{m + 3}{1 - 3m}$.Case $1$: $1 - 3m = m + 3 \implies 4m = -2 \implies m = -\frac{1}{2}$.Case $2$: $-(1 - 3m) = m + 3 \implies -1 + 3m = m + 3 \implies 2m = 4 \implies m = 2$.The lines are $y = -\frac{1}{2}x$ and $y = 2x$, which are $x + 2y = 0$ and $2x - y = 0$.The combined equation is $(x + 2y)(2x - y) = 0$.Expanding this: $2x^2 - xy + 4xy - 2y^2 = 0 \implies 2x^2 + 3xy - 2y^2 = 0$.

The combined equation of the pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with the line $3x + y - 6 = 0$ is...

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