Explanation

(a)
Given equation of the line is $x^{2} - 4xy + 3y^{2} = 0$
$\therefore\ \ \ m_{1} + m_{2} = \frac{4}{3}\ and\ m_{1}m_{2} = \frac{1}{3}$
On solving these equations, we get
$m_{1} = 1,\ m_{2} = \frac{1}{3}$
Let the lines parallel to given line are
$y = m_{1}x + c_{1}\$and$\ y = m_{2}x + c_{2}$
$\therefore y = \frac{1}{3}x + c_{1}\$and$\ y = x + c_{2}$
Also, these lines passes through the point $(3,\ - 2)$
$\therefore\ - 2 = \frac{1}{3} \times 3 + c_{1}\$
$\Rightarrow c_{1} = - 3$
$and - 2 = 1 \times 3 + c_{2}$
$\Rightarrow c_{2} = - 5$
$\therefore\$Required equation of pair of lines is
$(3y - x + 9)(y - x + 5) = 0$
$\Rightarrow \ x^{2} + 3y^{2} - 4xy - 14x + 24y + 45 = 0$