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

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$\frac{1}{2}$

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(A) Given relations are $l+m+n=0$ and $2lm+2nl-mn=0$.From the first equation,$l = -(m+n)$.Substituting this into the second equation:$2(-(m+n))m + 2n(-(m+n)) - mn = 0$$-2m^2 - 2mn - 2mn - 2n^2 - mn = 0$$-2m^2 - 5mn - 2n^2 = 0$$2m^2 + 5mn + 2n^2 = 0$$(2m+n)(m+2n) = 0$.Case $1$: $n = -2m$. Then $l = -(m - 2m) = m$. So $l=m$ and $n=-2m$. The direction ratios are $(1, 1, -2)$.Case $2$: $m = -2n$. Then $l = -(-2n + n) = n$. So $l=n$ and $m=-2n$. The direction ratios are $(1, -2, 1)$.Let the direction ratios be $\vec{a} = (1, 1, -2)$ and $\vec{b} = (1, -2, 1)$.The cosine of the angle $	heta$ is given by $\cos 	heta = \left| \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} ight|$.$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-2) + (-2)(1) = 1 - 2 - 2 = -3$.$|\vec{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$.$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.$\cos 	heta = \left| \frac{-3}{\sqrt{6} \cdot \sqrt{6}} ight| = \left| \frac{-3}{6} ight| = \frac{1}{2}$.

If $\theta$ is the acute angle between the two lines whose direction cosines are connected by the relations $l+m+n=0$ and $2lm+2nl-mn=0$,then $\cos \theta=$

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