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(A) Given equations for direction cosines are $a l + b m + c n = 0$ $(1)$ and $m n + n l + l m = 0$ $(2)$. From $(1)$, we have $n = -\frac{a l + b m}{

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

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(A) Given equations for direction cosines are $a l + b m + c n = 0$ $(1)$ and $m n + n l + l m = 0$ $(2)$. From $(1)$, we have $n = -\frac{a l + b m}{c}$. Substituting this into $(2)$: $m \left( -\frac{a l + b m}{c} ight) + l \left( -\frac{a l + b m}{c} ight) + l m = 0$. Multiplying by $-c$: $m(a l + b m) + l(a l + b m) - c l m = 0$. $a l^2 + b m^2 + a l m + b l m - c l m = 0$. $a l^2 + (a + b - c) l m + b m^2 = 0$. Dividing by $m^2$, we get $a \left( \frac{l}{m} ight)^2 + (a + b - c) \left( \frac{l}{m} ight) + b = 0$. Let the two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$. The roots of the quadratic are $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$. Thus, $\frac{l_1 l_2}{m_1 m_2} = \frac{b}{a}$, which implies $\frac{l_1 l_2}{b} = \frac{m_1 m_2}{a}$. By symmetry, $\frac{l_1 l_2}{1/a} = \frac{m_1 m_2}{1/b} = \frac{n_1 n_2}{1/c} = k$. The lines are perpendicular if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$. Substituting the values: $k \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ight) = 0$. Since $k 
eq 0$, the condition is $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$.

The lines whose direction cosines are given by the relations $a l+b m+c n=0$ and $m n+n l+l m=0$ are

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