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

We have,

$\begin{array}{r}a+b+c=0\text{ and }2ab+2ac-bc=0\\ \Rightarrow a=-(b+c)\text{ and }2a(b+c)-bc=0\\ \Rightarrow -2(b+c{)}^2-bc=0\\ \Rightarrow 2b^2+5bc+2c^2=0\\ \Rightarrow (2b+c)(b+2c)=0\\ \Rightarrow 2b+c=0\text{ or }b+2c=0\end{array}$

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

$\therefore a=b\text{ and }c=-2b\Rightarrow \frac{a}{1}=\frac{b}{1}=\frac{c}{-2}$

If $b+2c=0,\text{ then }a=-(b+c)\Rightarrow a=c$

$\therefore a=c\text{ and }b=-2c\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

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

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