Given direction cosines of two lines are
$$
\begin{aligned}
& l+m+n=0 \\
& \quad l^2-5 m^2+n^2=0
\end{aligned}
$$
and
Also,
$$
\begin{aligned}
l^2+m^2+n^2 & =1 \\
\left(I_1, m_1, n_1\right) & =\left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)
\end{aligned}
$$
$$
\text { and } \quad\left(I_2, m_2, n_2\right)=\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\right)
$$
$$
\begin{aligned}
\therefore & \cos \theta=\left|l_1 I_2+m_1 m_2+n_1 n_2\right| \\
= & \left|-\frac{2}{\sqrt{6}} \times \frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}} \times \frac{1}{\sqrt{6}}-\frac{2}{\sqrt{6}} \times \frac{1}{\sqrt{6}}\right| \\
\Rightarrow \cos \theta & =\left|-\frac{2}{6}+\frac{1}{6}-\frac{2}{6}\right|=\left|-\frac{3}{6}\right| \\
= & \frac{1}{2} \Rightarrow \theta=60^{\circ}=\frac{\pi}{3}
\end{aligned}
$$