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(C) The equation of a plane passing through the line of intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ is given by $

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$(ab' - a'b)y + (ac' - a'c)z + (ad' - a'd) = 0$

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(C) The equation of a plane passing through the line of intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ is given by $(ax + by + cz + d) + \lambda (a'x + b'y + c'z + d') = 0$. Rearranging the terms,we get $x(a + \lambda a') + y(b + \lambda b') + z(c + \lambda c') + (d + \lambda d') = 0$ ... $(i)$. Since this plane is parallel to the line $y = 0, z = 0$ (which is the $x$-axis),its normal vector must be perpendicular to the direction vector of the $x$-axis,which is $(1, 0, 0)$. Therefore,the dot product of the normal vector $(a + \lambda a', b + \lambda b', c + \lambda c')$ and $(1, 0, 0)$ must be zero: $1(a + \lambda a') + 0(b + \lambda b') + 0(c + \lambda c') = 0$. This gives $a + \lambda a' = 0$,so $\lambda = -\frac{a}{a'}$. Substituting $\lambda = -\frac{a}{a'}$ into equation $(i)$: $(ax + by + cz + d) - \frac{a}{a'}(a'x + b'y + c'z + d') = 0$. Multiplying by $a'$,we get $a'ax + a'by + a'cz + a'd - aa'x - ab'y - ac'z - ad' = 0$. Simplifying,we get $(a'b - ab')y + (a'c - ac')z + (a'd - ad') = 0$. Multiplying by $-1$,we obtain $(ab' - a'b)y + (ac' - a'c)z + (ad' - a'd) = 0$.

The equation of the plane passing 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

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