Given
$$
\begin{aligned}
&l+3 m+5 n=0 \\
&\text { and } 5 l m-2 m n+6 n l=0
\end{aligned}
$$
From eq. (1) we have
$$
l=-3 m-5 n
$$
Put the value of $l$ in eq. (2), we get;
$$
\begin{aligned}
&5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0 \\
&\Rightarrow 15 m^2+45 m n+30 n^2=0 \\
&\Rightarrow m^2+3 m n+2 n^2=0 \\
&\Rightarrow m^2+2 m n+m n+2 n^2=0 \\
&\Rightarrow(m+n)(m+2 n)=0 \\
&\therefore m=-n \text { or } m=-2 n \\
&\text { For } m=-n, l=-2 n \\
&\text { And for } m=-2 n, l=n
\end{aligned}
$$

$$
\begin{aligned}
&\therefore(l, m, n)=(-2 n,-n, n) \text { Or }(l, m, n) \\
&=(n,-2 n, n) \\
&\Rightarrow(l, m, n)=(-2,-1,1) \text { Or }(l, m, n) \\
&=(1,-2,1)
\end{aligned}
$$
Therefore, angle between the lines is given as:
$$
\begin{aligned}
&\cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}} \\
&\Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)
\end{aligned}
$$