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If the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies in the plane \(2 x-4 y+z=7\), then the value of \(k\) is
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The correct answer is:
7
The point \((4,2, k)\) on the line also lies on the plane \(2 x-4 y+z=7\)
So, \(8-8+\mathrm{k}=7 \Rightarrow \mathrm{k}=7\)
So, \(8-8+\mathrm{k}=7 \Rightarrow \mathrm{k}=7\)
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