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The equation of the plane through the line of intersection of the planes $ax+by+cz=d=0$ and $a'x+b'y+c'z+d'=0$ and parallel to the line $y=0,z=0$ is ?

$\begin{array}{1 1} (a)\:(ab'-a'b)x+(bc'-cb')y+(ad'-a'd)=0\:& \:(b)\:(ab'-a'b)x+(bc'-b'c)y+(ac'-a'c)z=0\\ (c)\:(ab'-a'b)y+(bc'-b'c)z+(ad'-a'd)=0\:& \:(d)none\:of\:these.\:\:\:\:\: \end{array} $

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Equation of any plane through the line of intersection of the given two planes
$ax+by+cz+d=0$ and $a'x+b'y+c'z+d'=0$ is given by
$\Rightarrow\:(a+\lambda a')x+(b+\lambda b')y+(c+\lambda c')z+(d+\lambda d')=0$
But given that this plane is parallel to the line $y=0,\:z=0$ (which is $x \:axis$).
$\Rightarrow\:$ The normal $\overrightarrow n=(a+\lambda a',b+\lambda b',c+\lambda c')$ is $\perp$ to the line.( i.e., $x \:axis.$)
$\Rightarrow\:(a+\lambda a',b+\lambda b',c+\lambda c').(1,0,0)=0$
$\Rightarrow\:a+\lambda a'=0\:\:\:or\:\:\:\lambda=-\large\frac{a}{a'}$
Substituting the value of $\lambda$ we get the required eqn. of the plane is
answered Jan 3, 2014 by rvidyagovindarajan_1

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