Explanation

(c)
Lines are $\left\lbrack \left( l + \sqrt{3}m \right)x + \left( m - \sqrt{3}l \right)y \right\rbrack\left\lbrack \left( l - \sqrt{3}m \right)x + \left( m + \sqrt{3}l \right)y \right\rbrack = 0$
and $lx + my + n = 0$
$\therefore m_{1} = - \frac{(l + \sqrt{3}m)}{(m - \sqrt{3}l)}\ ,\ m_{2} = - \frac{(l - \sqrt{3}m)}{(m + \sqrt{3}l)}$
$and\ m_{3} = - \frac{l}{m}$
$\therefore\ \theta_{1} = \tan^{- 1}\left\lbrack \frac{- \left( \frac{l + \sqrt{3}m}{m - \sqrt{3}l} \right) + \frac{l}{m}}{1 + \left( \frac{l + \sqrt{3}m}{m - \sqrt{3}l} \right) \bullet \frac{l}{m}} \right\rbrack = 60{^\circ}$
$and\ \theta_{2} = \tan^{- 1}\left\lbrack \frac{- \left( \frac{l - \sqrt{3}m}{m + \sqrt{3}l} \right) + \frac{l}{m}}{1 + \left( \frac{l - \sqrt{3}m}{m + \sqrt{3}l} \right)\left( \frac{l}{m} \right)} \right\rbrack = 60{^\circ}$
Hence, triangle is equilateral.