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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 this into the secon

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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 this into the second equation: $fm(-\frac{al + bm}{c}) + gl(-\frac{al + bm}{c}) + hlm = 0$.Multiplying by $-c$,we get $fm(al + bm) + gl(al + bm) - chlm = 0$.$aflm + bfm^2 + agl^2 + bglm - chlm = 0$.$agl^2 + (af + bg - ch)lm + bfm^2 = 0$.Dividing by $m^2$,we get $ag(\frac{l}{m})^2 + (af + bg - ch)(\frac{l}{m}) + bf = 0$.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 $l$,we get $\frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c}$.Thus,$\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c} = k$.For perpendicular lines,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.Substituting the values,$k(\frac{f}{a} + \frac{g}{b} + \frac{h}{c}) = 0$.Since $k 
eq 0$,we have $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

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

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