If the line frac{x - 4}{1} = frac{y - 2}{1} = | vedclass

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(B) The given line is $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$. The direction ratios of the line are $(1, 1, 2)$ and it passes through th

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$7$

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(B) The given line is $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$. The direction ratios of the line are $(1, 1, 2)$ and it passes through the point $(4, 2, k)$.The plane is $2x - 4y + z = 7$. The normal vector to the plane is $\vec{n} = (2, -4, 1)$.For the line to lie in the plane,the line must be parallel to the plane,meaning the dot product of the direction vector of the line and the normal vector of the plane must be zero:$(1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$.Since the dot product is $0$,the line is parallel to the plane.Additionally,the point $(4, 2, k)$ on the line must satisfy the equation of the plane for the line to lie entirely within it:$2(4) - 4(2) + k = 7$$8 - 8 + k = 7$$k = 7$.

If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2x - 4y + z = 7$,then $k = \dots$

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