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Condition for Coplanarity of Four Points

The equation ...

Question

The equation of line passing through $(2, 3, - 1)$ and lying in the plane $2 x + y - 5 z = 12$ and perpendicular to the line $2 x + y - 5 z - 12 = 0 = 4 x - 3 y + 7 z$ is :

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

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\frac{x - 1}{- 3} = \frac{y - 5}{1} = \frac{z + 1}{- 1}

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\frac{x - 5}{3} = \frac{y - 2}{- 1} = \frac{z}{1}

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\frac{x - 2}{- 3} = \frac{y - 2}{1} = \frac{z + 1}{7}

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Solution

The correct option is D $\frac{x - 2}{3} = \frac{y - 3}{- 1} = \frac{z + 1}{1}$
The equation of line passing through $(2, 3, - 1)$
and lying in the plane $2 x + y - 5 z = 12$
and perpendicular to the line $2 x + y - 5 z - 12 = 0 = 4 x - 3 y + 7 z$ is :
$\frac{x - 2}{3} = \frac{y - 3}{- 1} = \frac{z + 1}{1}$
Hence Option A is correct .

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Similar questions

Equation of the line passing through (1,1,1) and perpendicular to $2 x - 3 y + z = 0$ is

2 x - y + z = 3

4 x - 3 y + 5 z + 9 = 0

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π : 2 x - y + z - 4 = 0

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