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(C) Given equations for direction cosines $(l, m, n)$ are: $l+m+n=0$  --- $(i)$ $l^2+m^2-n^2=0$ --- $(ii)$ From $(i)$, $n = -(l+m)$. Substituting this

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$\frac{\pi}{3}$

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(C) Given equations for direction cosines $(l, m, n)$ are: $l+m+n=0$  --- $(i)$ $l^2+m^2-n^2=0$ --- $(ii)$ From $(i)$, $n = -(l+m)$. Substituting this into $(ii)$: $l^2 + m^2 - (-(l+m))^2 = 0$ $l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$ $-2lm = 0 \Rightarrow lm = 0$. This implies either $l=0$ or $m=0$. Case $1$: If $l=0$, then from $(i)$, $m+n=0 \Rightarrow m=-n$. Let $m=1$, then $n=-1$. The direction ratios are $(0, 1, -1)$. Case $2$: If $m=0$, then from $(i)$, $l+n=0 \Rightarrow l=-n$. Let $l=1$, then $n=-1$. The direction ratios are $(1, 0, -1)$. Let the two vectors be $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$. The angle $	heta$ between them is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$. $\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$. $|\vec{a}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$. $|\vec{b}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$. $\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$. Therefore, $	heta = \frac{\pi}{3}$.

If the direction cosines of two lines are such that $l+m+n=0$ and $l^2+m^2-n^2=0$, then the angle between them is

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