The angle between the lines whose direction cosines are given by the equations \(3 l+m+5 n=0\) and \(6 n m-2 n l+5 l m=0…

\(\text{Given one linear relation between direction cosines: } 3l + m + 5n = 0\)

\(\Rightarrow m = -3l - 5n\)

\(\text{Second relation: } 6nm - 2nl + 5lm = 0\)

\(\text{Substitute } m = -3l - 5n:\; 6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0\)

\(\Rightarrow -18ln - 30n^{2} - 2ln - 15l^{2} - 25ln = 0\)

\(\Rightarrow -15l^{2} - 45ln - 30n^{2} = 0\)

\(\Rightarrow l^{2} + 3ln + 2n^{2} = 0\)

\(\Rightarrow (l + n)(l + 2n) = 0\)

\(\text{Case 1: } l = -n \Rightarrow m = -3(-n) - 5n = -2n\)

\(\Rightarrow \text{direction ratios } \propto (l,m,n) = (-1,-2,1)\)

\(\text{Case 2: } l = -2n \Rightarrow m = -3(-2n) - 5n = n\)

\(\Rightarrow \text{direction ratios } \propto (l,m,n) = (-2,1,1)\)

\(\text{Angle } \theta \text{ between lines with ratios } \vec{a}=(-1,-2,1), \vec{b}=(-2,1,1)\)

\(\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\;\|\vec{b}\|} = \dfrac{(-1)(-2)+(-2)(1)+(1)(1)}{\sqrt{(-1)^2+(-2)^2+1^2}\;\sqrt{(-2)^2+1^2+1^2}}\)

\(\cos\theta = \dfrac{2 - 2 + 1}{\sqrt{6}\;\sqrt{6}} = \dfrac{1}{6}\)

\(\therefore \theta = \cos^{-1}\!\left(\dfrac{1}{6}\right)\)