Given, 3lm − 4ln + mn = 0 … (i)
l + 2m + 3n = 0 … (ii)
$\Rightarrow \quad l=-2 m-3 n$
From Eq. (i), we get
3(− 2m − 3n)m − 4(− 2m − 3n) n + mn = 0
$-6 m^2-9 n m+8 m n+12 n^2+m n=0$
$$
\begin{gathered}
\Rightarrow \quad 6 m^2=12 n^2 \Rightarrow m^2=2 n^2 \\
m= \pm \sqrt{2} n
\end{gathered}
$$
$\therefore\left(l_1, m_1, n_1\right)$ and $\left(l_2, m_2, n_2\right)$ are direction ratios where, $m_1=\sqrt{2} n_1$ and $m_2=\sqrt{2} n_2$
$\begin{aligned} l_1 & =-(2 \sqrt{2}+3) n_1 \text { and } l_2=-(-2 \sqrt{2}+3) n_2 \\ \therefore \cos \theta & =\frac{l_1 l_2+m_1 m_2+n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2} \sqrt{l_2^2+m_2^2+n_2^2}} \\ & =\frac{n_1 n_2[-2+(9-8)+1]}{\sqrt{l_1^2+m_1^2+n_1^2} \sqrt{l_2^2+m_2^2+n_2^2}}=0 \\ \cos \theta & =\cos \pi / 2 \Rightarrow \theta=\pi / 2\end{aligned}$