$l+3 m+5 n=0$ ....$(1)$
and $5 l m-2 m n+6 n l=0$ .....$(2)$
From eq. $( 1 )$ we have $l=-3 m-5 n$
Put the value of $l$ in eq. $(2),$ we get;
$5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0$
$\Rightarrow 15 m^{2}+45 m n+30 n^{2}=0$
$\Rightarrow m^{2}+3 m n+2 n^{2}=0$
$\Rightarrow m^{2}+2 m n+m n+2 n^{2}=0$
$\Rightarrow(m+n)(m+2 n)=0$
$\therefore m=-n$ or $m=-2 n$
For $m=-n, l=-2 n$
And for $m=-2 n, l=n$
$\therefore(l, m, n)=(-2 n,-n, n)$ Or $(l, m, n)$
$=(n,-2 n, n)$
$\Rightarrow(l, m, n)=(-2,-1,1)$ Or $(l, m, n)$
$=(1,-2,1)$
Therefore, angle between the lines is given as:
$\cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}}$
$\Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)$
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