Show that the straight lines whose direction cosines are given by $2l + 2m - n = 0$ and $mn + nl + lm = 0$ are at right angles.

It is given that, $2l + 2m - n = 0$

….(i)

and $mn + nl + lm = 0$

…..(ii)
Eliminating $m$ from the both equations,

we get
$m = \frac{{n - 2l}}{2}$

[from Eq. (i)]
$\Rightarrow$ $\left( {\frac{{n - 2l}}{2}} \right)n + nl + l\left( {\frac{{n - 2l}}{2}} \right) = 0$

$\Rightarrow$ $\frac{{{n^2} - 2nl + 2nl + nl - 2{l^2}}}{2} = 0$
$\Rightarrow$ ${n^2} + nl - 2{l^2} = 0$
$\Rightarrow$ ${n^2} + 2nl - nl - 2{l^2} = 0$

$\Rightarrow$ $(n + 2l)(n - l) = 0$
$\Rightarrow$ $n = - 2l$ and $n = l$
$\therefore m = \frac{{ - 2l - 2l}}{2},m = \frac{{l - 2l}}{2}$

$\Rightarrow$ $m = - 2l,m = \frac{{ - l}}{2}$

Thus, the direction ratios of two lines are proportional to $l, - 2l, - 2$ and $l,\frac{{ - l}}{2},l$.

$\Rightarrow$ $1, - 2, - 2$ and $1,\frac{{ - 1}}{2},1$

$\Rightarrow$ $1, - 2, - 2$ and 2,-1,2
Also,

the vectors parallel to these lines are $\overrightarrow {\rm{a}} = \widehat {\rm{i}} - 2\widehat {\rm{j}} - 2\widehat {\rm{k}}$ and $\overrightarrow {\rm{b}}$

$= 2\widehat {\rm{i}} - \widehat {\rm{j}} + 2\widehat {\rm{k}}$ respectively.

$\therefore \cos \theta = \frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{a}} |\overrightarrow {{\rm{b}}\mid } }}$

$= \frac{{(\widehat {\rm{i}} - 2\widehat {\rm{j}} - 2\widehat {\rm{k}}) \cdot (2\widehat {\rm{i}} - \widehat {\rm{j}} + 2\widehat {\rm{k}})}}{{3 \cdot 3}}$

$= \frac{{2 + 2 - 4}}{9} = 0$
$\therefore \theta = \frac{\pi }{2}$