Find the angle between the lines whose direction cosines are given by the equations $3l+m+5n=0$ and $6mn-2nl+5lm=0$.

(A) Given equations are $3l+m+5n=0$ $(1)$ and $6mn-2nl+5lm=0$ $(2)$.
From $(1)$,$m = -(3l+5n)$.
Substituting this into $(2)$: $6n(-(3l+5n)) - 2nl + 5l(-(3l+5n)) = 0$.
$-18ln - 30n^2 - 2nl - 15l^2 - 25ln = 0$.
$-15l^2 - 45ln - 30n^2 = 0$.
Dividing by $-15$: $l^2 + 3ln + 2n^2 = 0$.
$(l+n)(l+2n) = 0$.
Case $1$: $l = -n$. Then $m = -(3(-n)+5n) = -2n$. Direction ratios are $(-n, -2n, n)$,i.e.,$(1, 2, -1)$.
Case $2$: $l = -2n$. Then $m = -(3(-2n)+5n) = n$. Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$.
Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} + \hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|(1)(-2) + (2)(1) + (-1)(1)|}{\sqrt{1^2+2^2+(-1)^2} \sqrt{(-2)^2+1^2+1^2}} = \frac{|-2+2-1|}{\sqrt{6}\sqrt{6}} = \frac{1}{6}$.
Thus,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.