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(C) Let the equation of the family of lines passing through the intersection be $(ax + 2by + 3b) + \lambda(bx - 2ay - 3a) = 0$. Since the line is para

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below $x$-axis at a distance $\frac{3}{2}$ from it

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(C) Let the equation of the family of lines passing through the intersection be $(ax + 2by + 3b) + \lambda(bx - 2ay - 3a) = 0$. Since the line is parallel to the $x$-axis,its slope must be $0$. The equation can be rewritten as $(a + \lambda b)x + (2b - 2a\lambda)y + (3b - 3a\lambda) = 0$. For the line to be parallel to the $x$-axis,the coefficient of $x$ must be $0$,so $a + \lambda b = 0$,which gives $\lambda = -\frac{a}{b}$. Substituting $\lambda = -\frac{a}{b}$ into the equation: $(ax + 2by + 3b) - \frac{a}{b}(bx - 2ay - 3a) = 0$ $ax + 2by + 3b - ax + \frac{2a^2}{b}y + \frac{3a^2}{b} = 0$ $(2b + \frac{2a^2}{b})y = -(\frac{3a^2}{b} + 3b)$ $(\frac{2b^2 + 2a^2}{b})y = -\frac{3(a^2 + b^2)}{b}$ $2(a^2 + b^2)y = -3(a^2 + b^2)$ Since $(a, b) 
eq (0, 0)$,$a^2 + b^2 
eq 0$,so we can divide by $a^2 + b^2$: $2y = -3 \Rightarrow y = -\frac{3}{2}$. This represents a line parallel to the $x$-axis at a distance of $\frac{3}{2}$ below the $x$-axis.

The line parallel to the $x$-axis passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$,where $(a, b) \neq (0, 0)$,is

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