$\overline{ r }=|\overline{ r }|\left(\pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{ j } \pm \frac{1}{\sqrt{3}} \hat{ k }\right)$
For equally inclined to co-ordinate axes.
$\alpha=\beta=\gamma$ and
$1= m = n$
Since, $1^{2}+m^{2}+n^{2}=1$
$\Rightarrow 31^{2}=1$
$\Rightarrow 1^{2}=\frac{1}{3}$
$\therefore l =\pm \frac{1}{\sqrt{3}}= m = n$
$ \Rightarrow 1, m, n $ each has $2$ choices.
$\therefore $ total lines $=2^{3}$