$2 l-m+3 n=0$ [Given]
$$
m=2 l+3 n
$$
Substitute this value of $m$ in $l^2+m n-6 n^2=0$
$$
\begin{aligned}
& l^2+(2 l+3 n) n-6 n^2=0 \\
\Rightarrow & l^2+2 l n-3 n^2=0 \\
\Rightarrow & l=\frac{-2 n \pm \sqrt{4 n^2+12 n^2}}{2} \\
\Rightarrow & l=\frac{-2 n+4 n}{2}, \frac{-2 n-4 n}{2} \\
\Rightarrow & l=n,-3 n
\end{aligned}
$$
Substitute these value in $m=2 l+3 n$
$$
m=5 n,-3 n
$$
DR's are $(n, 5 n, n)$ and $(-3 n,-3 n, n)$.
DC's are $\left(\frac{1}{3 \sqrt{3}}, \frac{5}{3 \sqrt{3}}, \frac{1}{3 \sqrt{3}}\right)$ and $\left(\frac{-3}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{1}{\sqrt{19}}\right)$
Here $l_1=\frac{1}{3 \sqrt{3}}, l_2=\frac{-3}{\sqrt{19}}, m_1=\frac{5}{3 \sqrt{3}}, m_2=\frac{-3}{\sqrt{19}}$
$$
\begin{aligned}
&\left|l_1 l_2\right|+\left|m_1 m_2\right| \\
&=\left|\frac{-1}{\sqrt{3} \sqrt{19}}\right|+\left|\frac{-5}{\sqrt{3} \sqrt{19}}\right| \\
& \quad=\frac{6}{\sqrt{3} \sqrt{19}}=\frac{2 \sqrt{3}}{\sqrt{19}}
\end{aligned}
$$