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(A) Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$. From the first equation,$n = -\frac{al + bm}{c}$. Substituting $n$ in the second

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$\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$

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(A) Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$. From the first equation,$n = -\frac{al + bm}{c}$. Substituting $n$ in the second equation: $fm(-\frac{al + bm}{c}) + gl(-\frac{al + bm}{c}) + hlm = 0$. Multiplying by $-c$,we get $fm(al + bm) + gl(al + bm) - chlm = 0$. Expanding: $aflm + bfm^2 + agl^2 + bglm - chlm = 0$. Rearranging as a quadratic in $(l/m)$: $ag(\frac{l}{m})^2 + (af + bg - ch)(\frac{l}{m}) + bf = 0$. Let the roots be $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$. Then $\frac{l_1 l_2}{m_1 m_2} = \frac{bf}{ag}$,which implies $\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b}$. Similarly,by eliminating $l$ and $m$,we get $\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c} = k$. For the lines to be perpendicular,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$. Substituting the ratios: $k(\frac{f}{a} + \frac{g}{b} + \frac{h}{c}) = 0$. Since $k 
eq 0$,we must have $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular if:

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