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

(A) We have,$2l + 2m - n = 0 \dots (i)$
And $mn + nl + lm = 0 \dots (ii)$
From $(i)$,$n = 2l + 2m$. Substituting this into $(ii)$:
$m(2l + 2m) + (2l + 2m)l + lm = 0$
$2lm + 2m^2 + 2l^2 + 2lm + lm = 0$
$2l^2 + 5lm + 2m^2 = 0$
$2l^2 + 4lm + lm + 2m^2 = 0$
$2l(l + 2m) + m(l + 2m) = 0$
$(2l + m)(l + 2m) = 0$
Case $1$: $m = -2l$. Then $n = 2l + 2(-2l) = -2l$. The direction ratios are $(l, -2l, -2l)$,which are proportional to $(1, -2, -2)$.
Case $2$: $l = -2m$. Then $n = 2(-2m) + 2m = -2m$. The direction ratios are $(-2m, m, -2m)$,which are proportional to $(2, -1, 2)$.
Let the direction ratios be $a_1, b_1, c_1 = (1, -2, -2)$ and $a_2, b_2, c_2 = (2, -1, 2)$.
The condition for perpendicularity is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
$(1)(2) + (-2)(-1) + (-2)(2) = 2 + 2 - 4 = 0$.
Since the sum is $0$,the lines are at right angles.