# The value of k such that x-4 /1 = y-2/1 = z-k / 2 lies in the plane 2x-4y+z=7 is

Solution:   Given  ,

$L_{1}: \frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2} \hspace{0.5cm}P: 2x-4y+z=7$

The point   $(4,2,k)$  through  which it passes must also  lie on the

Given plane ,

Hence ,     $2\times4-4\times 2+k=7\Rightarrow k=7.$

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