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(C) Given equations are $3l + m + 5n = 0$ and $6mn - 2nl + 5lm = 0$. From the first equation, $m = -3l - 5n$. Substituting this into the second equati

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$\frac{1}{6}$

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(C) Given equations are $3l + m + 5n = 0$ and $6mn - 2nl + 5lm = 0$. From the first equation, $m = -3l - 5n$. Substituting this into the second equation: $6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0$ $-18nl - 30n^2 - 2nl - 15l^2 - 25nl = 0$ $-15l^2 - 45nl - 30n^2 = 0$ Dividing by $-15$: $l^2 + 3nl + 2n^2 = 0$ $(l + n)(l + 2n) = 0$ This gives two cases: Case $1$: $l = -n$. Substituting into $m = -3l - 5n$, we get $m = -3(-n) - 5n = 3n - 5n = -2n$. The direction ratios are $(-n, -2n, n)$, which simplifies to $(1, 2, -1)$. Case $2$: $l = -2n$. Substituting into $m = -3l - 5n$, we get $m = -3(-2n) - 5n = 6n - 5n = n$. The direction ratios are $(-2n, n, n)$, which simplifies to $(-2, 1, 1)$. Let the direction ratios be $\vec{a} = (1, 2, -1)$ and $\vec{b} = (-2, 1, 1)$. The cosine of the angle $	heta$ is given by: $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(-2) + (2)(1) + (-1)(1)|}{\sqrt{1^2 + 2^2 + (-1)^2} \sqrt{(-2)^2 + 1^2 + 1^2}}$ $\cos 	heta = \frac{|-2 + 2 - 1|}{\sqrt{6} \sqrt{6}} = \frac{|-1|}{6} = \frac{1}{6}$.

The direction cosines $l, m, n$ of two lines satisfy the equations $3l + m + 5n = 0$ and $6mn - 2nl + 5lm = 0$. If $\theta$ is the angle between these lines, then $|\cos \theta| = $

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