Explanation

(c)
$3l + m + 5n = 0$ (i)
$6mn - 2nl + 5ml = 0$ (ii)
Substituting the value of $n$ from Eq. (i) in Eq. (ii), we get
$6l^{2} + 9lm - 6m^{2} = 0$
$\Rightarrow \ 6\left( \frac{l}{m} \right)^{2} + 9\left( \frac{l}{m} \right) - 6 = 0$
$\therefore\frac{l_{1}}{m_{1}} = \frac{1}{2}\$ and $\frac{l_{2}}{m_{2}} = - 2$
From Eq. (i), we get
$\frac{l_{1}}{n_{1}} = - 1\$ and $\frac{l_{2}}{n_{2}} = - 2$
$\therefore\frac{l_{1}}{1} = \frac{m_{1}}{2} = \frac{n_{1}}{- 1} = \sqrt{\frac{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}}{1 + 4 + 1}} = \frac{1}{\sqrt{6}}$
and $\frac{l_{2}}{2} = \frac{m_{2}}{- 1} = \frac{n_{2}}{- 1} = \frac{\sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}{\sqrt{4 + 1 + 1}} = \frac{1}{\sqrt{6}}$
If $\theta$ be the angle between the lines, then
$\cos\theta = \left( \frac{1}{\sqrt{6}} \right)\left( \frac{2}{\sqrt{6}} \right) + \left( \frac{2}{\sqrt{6}} \right)\left( - \frac{1}{\sqrt{6}} \right) + \left( - \frac{1}{\sqrt{6}} \right)\left( - \frac{1}{\sqrt{6}} \right) = \frac{1}{6}$
$\therefore\ \theta = \cos^{- 1}\left( \frac{1}{6} \right)$