Explanation

(d)
Here, $a = 6,\ 2h = - 1,\ b = 4c$
$\therefore m_{1} + m_{2} = \frac{1}{4c},\ m_{1}m_{2} = \frac{6}{4c}$
One line of given pair of line is $3x + 4y = 0$
$Slope\ of\ line = - \frac{3}{4} = m_{1}(say)$
$\therefore\ - \frac{3}{4} + m_{2} = \frac{1}{4c}$
$\Rightarrow \ m_{2} = \frac{1}{4c} + \frac{3}{4}$
$\therefore\ \left( - \frac{3}{4} \right)\left( \frac{1}{4c} + \frac{3}{4} \right) = \frac{6}{4c}$
$\Rightarrow \ 1 + 3c = \frac{- 6 \times 4}{3}$
$\Rightarrow \ 3c = - 9\ \ \Rightarrow \ c = - 3$