The angle between the lines whose direction cosines are given by...

The given equations are $3 l+m+5 n=0$...(i)$and 6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{m}=0$...(ii) From (i), we have $m=-3 l-5 n$. Putting $\mathrm{m}=-3 l-5 \mathrm{n}$ in (ii), we get $6(-3 l-5 n) n-2 n l+5 l(-3 l-5 n)=0\Rightarrow(n+l)(2 n+l)=0\Rightarrow$ either $l=-\mathrm{n}$ or $l=-2 \mathrm{n}$. If $l=-n$, then putting $l=-n$ in (i), we obtain $m=-2 n$. If $l=-2 \mathrm{n}$, then putting $l=-2 \mathrm{n}$ in (i), we obtain $\mathrm{m}=\mathrm{n}$. Thus, the direction ratios of two lines are $-\mathrm{n},-2 \mathrm{n}, \mathrm{n} and -2 \mathrm{n}, \mathrm{n}, \mathrm{n}$ i.e., $1,2,-1 and -2,1,1$. Hence, the direction cosines are $\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}$ or $\frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}$. The angle $\theta$ between the lines is given by $$\begin{aligned} & \cos \theta=\frac{1}{\sqrt{6}} \times \frac{-2}{\sqrt{6}}+\frac{2}{\sqrt{6}} \times \frac{1}{\sqrt{6}}+\frac{-1}{\sqrt{6}} \times \frac{1}{\sqrt{6}}=\frac{-1}{6} \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{-1}{6}\right) \end{aligned}$$