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(B) Given relations are $l + m + n = 0$ and $2lm + 2nl - mn = 0$. From the first equation,$n = -(l + m)$. Substituting this into the second equation:

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

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(B) Given relations are $l + m + n = 0$ and $2lm + 2nl - mn = 0$. From the first equation,$n = -(l + m)$. Substituting this into the second equation: $2lm + 2(l + m)(-(l + m)) - m(-(l + m)) = 0$. $2lm - 2(l^2 + 2lm + m^2) + m(l + m) = 0$. $2lm - 2l^2 - 4lm - 2m^2 + ml + m^2 = 0$. $-2l^2 - lm - m^2 = 0$,which simplifies to $2l^2 + lm + m^2 = 0$. Wait,let us re-evaluate: $2lm + 2nl - mn = 0$. Substituting $n = -(l + m)$: $2lm + 2l(-(l + m)) - m(-(l + m)) = 0$. $2lm - 2l^2 - 2lm + ml + m^2 = 0$. $-2l^2 + lm + m^2 = 0$,or $2l^2 - lm - m^2 = 0$. Factoring gives $(2l + m)(l - m) = 0$. Case $1$: $2l + m = 0 \Rightarrow m = -2l$. Then $n = -(l - 2l) = l$. Direction ratios $(l, m, n)$ are $(l, -2l, l)$,i.e.,$(1, -2, 1)$. Case $2$: $l - m = 0 \Rightarrow m = l$. Then $n = -(l + l) = -2l$. Direction ratios $(l, m, n)$ are $(l, l, -2l)$,i.e.,$(1, 1, -2)$. Let $	heta$ be the angle between the lines. $\cos 	heta = \frac{|(1)(1) + (-2)(1) + (1)(-2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{1^2 + 1^2 + (-2)^2}} = \frac{|1 - 2 - 2|}{\sqrt{6} \sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$. Since the angle between lines is usually taken as acute,$\cos 	heta = 1/2 \Rightarrow 	heta = \pi/3$. However,if we consider the angle between the vectors,$\cos 	heta = \frac{-3}{6} = -1/2 \Rightarrow 	heta = 2\pi/3$.

The angle between the lines whose direction cosines are connected by the relations $l + m + n = 0$ and $2lm + 2nl - mn = 0$ is

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