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(A) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.This can be factored as $(x - 2y)(x - 2y + k) = 0$.Thus, the two lines are $L_1: x - 2y = 0$ an

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$x - 2y = 0$ and $x - 2y + 3\sqrt{5} = 0$

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(A) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.This can be factored as $(x - 2y)(x - 2y + k) = 0$.Thus, the two lines are $L_1: x - 2y = 0$ and $L_2: x - 2y + k = 0$.The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Here, $a = 1$, $b = -2$, $c_1 = 0$, and $c_2 = k$.Given $d = 3$, we have $3 = \frac{|0 - k|}{\sqrt{1^2 + (-2)^2}}$.$3 = \frac{|-k|}{\sqrt{1 + 4}} = \frac{k}{\sqrt{5}}$ (since $k > 0$).$k = 3\sqrt{5}$.Substituting $k$ back into the equation for $L_2$, we get $x - 2y + 3\sqrt{5} = 0$.Therefore, the lines are $x - 2y = 0$ and $x - 2y + 3\sqrt{5} = 0$.

If the distance between the lines represented by $(x - 2y)^2 + k(x - 2y) = 0$ is $3$ units and $k > 0$, then the equations of the lines are:

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