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

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Condition for Two Lines to Be Perpendicular

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The value of k such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane 2x -4y +z =7, is

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The value of
$k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$
lies in the plane $2 x - 4 y + z = 7$ is

k

\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}

2 x - 4 y + z = 7

\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}

2 x - 4 y + z = 7

k

^{'} k^{'}

\frac{x - 3}{3} = \frac{y - 1}{- 2} = \frac{z - k}{- 8}

2 x - y + z = 10,

L_{1} : \frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}

2 x - 4 y + z = 7

L_{1}

2 x - 4 y + z = 7,

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