If the direction cosines of two lines are given by $l+m+n=0$ and $l^2-5m^2+n^2=0$ then the angle between them is 

$l+m+n=0\Rightarrow l=-m-n$

$l^2-5m^2+n^2=0\Rightarrow {\left(-m-n\right)}^2-5m^2+n^2=0\Rightarrow m^2+n^2+2mn-5m^2+n^2=0$

$\Rightarrow -4m^2+2n^2+2mn=0$$\Rightarrow 2m^2-n^2-mn=0\Rightarrow \left(m-n\right)\left(2m+n\right)=0\Rightarrow m=n or m=-n/2$

If $m=n$ then $l=-n-n=-2n\Rightarrow l:m:n=-2n:n:n=-2:1:1$

If $m=-n/2$ then $l=n/2-n=-n/2\Rightarrow l : m : n=-n/2:-n/2 :n=1 :1 :-2$

If $\theta$ is the angle between the lines then $\cos \theta =\left|\frac{\left(-2\right)\left(1\right)+\left(1\right)\left(1\right)+\left(1\right)\left(-2\right)}{\sqrt{4+1+1}\sqrt{1+1+4}}\right|=\frac{3}{6}=\frac{1}{2}\Rightarrow \theta =\frac{\mathrm{\pi }}{3}.$