We have,
and
$$
l+3 m+5 n=0
$$
and
$$
5 m-2 m n+6 l n=0
$$
On substituting the value of $l$ from Eq. (i) in Eq. (ii), we get
$$
\begin{array}{rcr}
& 5(-3 m-5 n) m-2 m n+6(-3 m-5 n) n=0 \\
\Rightarrow & -15 m^2-25 m n-2 m n-18 m n-30 n^2=0 \\
\Rightarrow & 15 m^2+45 m n+30 n^2=0 \\
\Rightarrow & m^2+3 m n+2 n^2=0 \\
\Rightarrow & (m+n)(m+2 n)=0 \\
\Rightarrow & m=-n \text { or } m=-2 n
\end{array}
$$
When $m=-n_{\text {r }}$ then $l=-2 n$ and thus the DR'S are 2. $-1,1$.
When $m=-2 n$, then $l=n$ and thus the DR'S are $1,1,-2$.
Hence, the DR's of given lines are $2,-1,1$ and $1,1,-2$.

Now, let $\theta$ be the acute angle between the given Ines, then
$$
\begin{aligned}
\cos \theta & =\frac{\mid 2 \cdot 1+(-1) \cdot 1+1 \cdot(-2 \mid}{\sqrt{2^2+(-1)^2+(1)^2} \sqrt{1^2+1^2+(-2)^2}} \\
\Rightarrow \quad \cos \theta & =\frac{1}{\sqrt{6} \sqrt{6}}=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)
\end{aligned}
$$