Given $2 l+2 m-n=0$
and $\mathrm{mn}+\mathrm{nl}+\mathrm{lm}=0$
$\mathrm{m}=\frac{\mathrm{n}-21}{2} \quad$ [from Eq. (i)]
$\Rightarrow\left(\frac{\mathrm{n}-21}{2}\right) \mathrm{n}+\mathrm{nl}+1\left(\frac{\mathrm{n}-21}{2}\right)=0$
$\Rightarrow \quad \frac{\mathrm{n}^2-2 \mathrm{nl}+2 \mathrm{nl}+\mathrm{nl}-2 \mathrm{l}^2}{2}=0$
$\Rightarrow \mathrm{n}^2+\mathrm{nl}-2 \mathrm{l}^2=0$
$\Rightarrow \quad(\mathrm{n}+21)(\mathrm{n}-1)=0$
$\Rightarrow \mathrm{n}=-21$ and $\mathrm{n}=1$

$$
\begin{aligned}
&\therefore \quad \mathrm{m}=\frac{-21-21}{2}, \mathrm{~m}=\frac{1-21}{2} \\
&\Rightarrow \quad \mathrm{m}=-21, \mathrm{~m}=\frac{-1}{2}
\end{aligned}
$$
Thus, the direction ratios of two lines are proportional to $1,-21,-2$ and $1, \frac{-1}{2}, 1$
$\Rightarrow 1,-2,-2$ and $1, \frac{-1}{2}, 1$
$\Rightarrow 1,-2,-2$ and $2,-1,2$
Also, the vectors parallel to these lines are $\vec{a}=\hat{i}-2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}$ respectively.
$$
\begin{aligned}
&\therefore \quad \cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}|}=\frac{(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{3 \cdot 3} \\
&=\frac{2+2-4}{9}=0
\end{aligned}
$$
$\therefore \quad \theta=\frac{\pi}{2} \quad\left[\because \cos \frac{\pi}{2}=0\right]$