Given direction cosines are $al+bm+cn=0 ...\left(1\right)$

$\Rightarrow n=-\left(\frac{al+bm}{c}\right) ...\left(2\right)$

and $mn+nl+lm=0 ...\left(3\right)$

From equation $\left(1\right)$,

$-m\left(\frac{al+bm}{c}\right)-l\left(\frac{al+bm}{c}\right)+lm=0$

$\Rightarrow m(al+bm)+l(al+bm)-lcm=0$

$\Rightarrow a{l}^2+lm\left(a+b-c\right)+b{m}^2=0$

Dividing by $m^2$,

$\Rightarrow \frac{a{l}^2}{m^2}+\frac{l}{m}\left(a+b-c\right)+b=0$

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of two lines,

then $\frac{l_1}{m_1}, \frac{l_1}{m_2}$ are roots of equation.

$\Rightarrow$ Product of roots, $\frac{l_1{l}_2}{m_1{m}_2}=\frac{b}{a}$

$\Rightarrow \frac{l_1{l}_2}{\left({\displaystyle \frac{1}{a}}\right)}=\frac{m_1{m}_2}{\left({\displaystyle \frac{1}{b}}\right)}=\frac{n_1{n}_2}{\left({\displaystyle \frac{1}{c}}\right)}=k$

The lines are perpendicular if

$l_1{l}_2+m_1{m}_2+n_1{n}_2=0$

$\Rightarrow k\left(\frac{1}{a}\right)+k\left(\frac{1}{b}\right)+k\left(\frac{1}{c}\right)=0$

$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$