Explanation

(b)
Eliminating $n$, we get
$\lambda(l + m)^{2} + lm = 0$
$\Rightarrow \frac{\lambda l^{2}}{m^{2}} + (2\lambda + 1)\frac{l}{m} + \lambda = 0$
$\Rightarrow \frac{l_{1}l_{2}}{m_{1}m_{2}} = 1$ (product of roots $\frac{l_{1}}{m_{1}}$ and $\frac{l_{2}}{m_{2}}$)
Where $l_{1}/m_{1}$ and $l_{2}/m_{2}$ are the roots of this equation, further eliminating $m$, we get
$\lambda l^{2} - \ln - n^{2} = 0$
$\Rightarrow \frac{l_{1}l_{2}}{n_{1}n_{2}} = - \frac{1}{\lambda}$
Since the lines with direction cosines $(l_{1},\ m_{1},\ n_{1})$ and $(l_{2},m_{2},\ n_{2})$ are perpendicular, we have
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0$
$\Rightarrow 1 + 1 - \lambda = 0$
$\Rightarrow \ \lambda = 2$