The angle between the lines whose direction cosines are given by

 the equations $l^2+m^2-n^2=0$  , $l+m+n=0$ is 

We are given  $l^2+m^2-n^2=0$ and $l+m+n=0$ also we have  $l^2+m^2+n^2=1$
So that   $2n^2=1\Rightarrow n=\pm \frac{1}{\sqrt{2}}$ and  $l+m=-n$
$\Rightarrow {\left(l+m\right)}^2=n^2=l^2+m^2\Rightarrow 2lm=0$
$\Rightarrow \text{Either\hspace{0.17em}\hspace{0.17em}}l=0orm=0\text{\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}\hspace{0.17em}}l=0,m+n=0$
$\Rightarrow m=-n=\pm \frac{1}{\sqrt{2}}$
So the direction cosines of one of the lines are, $0,\pm \frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{2}}$ and if  $m=0,l+n=0\Rightarrow l=-n$ $=\pm \frac{1}{\sqrt{2}}$  and the direction cosines of the other line are   $\pm \frac{1}{\sqrt{2}},0,\pm \frac{1}{\sqrt{2}}.$
Hence the required angle is  $\frac{\pi }{3}$