If the direction ratios of two lines are given by a + b + c = 0 and 2ab + 2 ac - bc = 0, then the angle between the lines, is

$\frac{2\pi}{3}$

We have,

$a + b + c= 0 $ and $ 2ab + 2ac - bc = 0.$

$⇒ a = -(b + c) $ and $ 2a (b + c) - bc = 0 $

$⇒ -2 ( b + c)^2 - bc = 0 $

$⇒ 2b^2 + 5bc = 2c^2 = 0 $

$⇒ (2b +c) (b + 2c) = 0 $

$⇒ 2b + c = 0 $ or, $ b + 2c = 0 $

 If $ 2b + c = 0 $, then $ a = - (b + c) ⇒ a = b $

$∴ a = b $ and $ c = - 2b ⇒ \frac{a}{1}=\frac{b}{1}=\frac{c}{-2}$

If $b + 2c = 0 $ , then $ a= -(b + c) ⇒ a = c $

$∴ a= c $ and $ b = - 2 c ⇒ \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

Thus, the direction ratios of two lines are proportional to 1, 1, -2 and 1, -2, 1 respectively. SO, the angle $\theta $ between them is given by 

$cos \theta = \frac{1-2-2}{\sqrt{1+1+4}\sqrt{1+4+1}}=\frac{-1}{2}⇒ \theta = \frac{2\pi}{3}$