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(B) The equation of any plane passing through the given line is $(x + y + 2z - 3) + \lambda(2x + 3y + 4z - 4) = 0$.Rearranging the terms,we get $(1 +

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$2$

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(B) The equation of any plane passing through the given line is $(x + y + 2z - 3) + \lambda(2x + 3y + 4z - 4) = 0$.Rearranging the terms,we get $(1 + 2\lambda)x + (1 + 3\lambda)y + (2 + 4\lambda)z - (3 + 4\lambda) = 0$.If this plane is parallel to the $z$-axis,its normal vector $\vec{n} = (1 + 2\lambda, 1 + 3\lambda, 2 + 4\lambda)$ must be perpendicular to the $z$-axis (which has direction vector $\vec{k} = (0, 0, 1)$).Therefore,$(1 + 2\lambda)(0) + (1 + 3\lambda)(0) + (2 + 4\lambda)(1) = 0$.This gives $2 + 4\lambda = 0$,so $\lambda = -\frac{1}{2}$.Substituting $\lambda = -\frac{1}{2}$ into the plane equation:$(x + y + 2z - 3) - \frac{1}{2}(2x + 3y + 4z - 4) = 0$.Multiplying by $2$: $2x + 2y + 4z - 6 - 2x - 3y - 4z + 4 = 0$,which simplifies to $-y - 2 = 0$,or $y + 2 = 0$.The $z$-axis is the line $x = 0, y = 0$. The distance from any point on the $z$-axis (e.g.,$(0, 0, 0)$) to the plane $y + 2 = 0$ is given by $d = \frac{|0 + 2|}{\sqrt{0^2 + 1^2 + 0^2}} = \frac{2}{1} = 2$.

The shortest distance between the $z$-axis and the line $x + y + 2z - 3 = 0 = 2x + 3y + 4z - 4$ is

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