The angle between the lines whose direction cosines satisfy the equations $l+m+n=0\text{ and }{l}^2=m^2-n^2$ is

We have,$I+ m + n = 0$

and, $l^2=m^2+n^2$

$\begin{array}{r}\therefore (-m-n{)}^2=m^2+n^2\\ \Rightarrow 2mn=0\Rightarrow m=0\text{ or }n=0\end{array}$

Now,

$\begin{array}{r}m=0\Rightarrow l+n=0\text{ and }{l}^2=n^2[\text{ Putting }m=0\text{ in (i) \& (ii) }]\\ \Rightarrow l=-n\end{array}$

Thus, the direction ratios of one of the two lines are proportional to

$-n,0,n\text{ or }-1,0,1$

When$n =0,$

$l+m+n=0$ and $l^2=m^2+n^2$

$\Rightarrow l+m=0\text{ and }{l}^2=m^2\Rightarrow l=-m$

Thus, the direction ratios of one of the two lines are proportional to -$m, m,$ 0 or -1, 1, 0. Let 0 be the angle between the given lines. Then, 

$\cos \theta =\frac{-1\times (-1)+0\times 1+1\times 0}{\sqrt{(-1{)}^2+0^2+1^2}\sqrt{(-1{)}^2+1^2+0}}=\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$