If the direction cosines of two lines are connected by the equations $ l + m + n = 0 , l^2 + m^2 - n^2 = 0 , $ then the angle between the lines, is

$\frac{\pi}{3}$

We have, 

$l + m + n = 0 $ and $ l^2 + m^2 - n^2 = 0 $

$⇒ l^2 + m^2 - (-l - m)^2 = 0 ⇒ ln = 0 ⇒ l = 0$ or $ m = 0 $

When $ l = 0 $:

$ l + m + n = 0 $ and $ l^2 + m^2 - n^2 = 0 $

$ ⇒ m +  n = 0 $ and $ m^2- n^2 = 0 ⇒ m + n = 0 ⇒ m = -n $

So, the direction ratios are proportional to $: \frac{l}{0}=\frac{m}{1}=\frac{n}{-1}$

When $ m = 0 $:

$ l + m + n = 0 $ and $ l^2 + m^2 - n^2 = 0 $

$ l + n = 0 $ and $ l^2 - n^2 = 0 ⇒ l + n = 0 ⇒ l = -n $

So, the direction ratio are proportional to : $\frac{l}{1}=\frac{m}{0}=\frac{n}{-1}$

Let $\theta $ be the angle between the lines. Then,

$ cos \theta =\frac{1×0+1×0+(-1)×(-1)}{\sqrt{0+1+1}\sqrt{1+0+1}}=\frac{1}{2}⇒ \frac{\pi}{3}$