Equation of the line passing through $(1, 1, 1)$ and parallel to the plane $2 x + 3 y + z + 5 = 0$ is

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

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\frac{x - 1}{- 1} = \frac{y - 1}{1} = \frac{z - 1}{- 1}

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

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

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Solution

The correct option is A $\frac{x - 1}{- 1} = \frac{y - 1}{1} = \frac{z - 1}{- 1}$
Line parallel to $2 x + 3 y + z + 5 = 0$ means it is perpendicular to normal to plane.
If direction cosines of line are $α, β$ and $γ$ then
$2 α + 3 β + γ = 0$
It is satisfied if $α = - 1, β = 1, γ = - 1$
$∴$ Equation of line passing through $(1, 1, 1)$
having D.C's $(- 1, 1, - 1)$ is
$\frac{x - 1}{- 1} = \frac{y - 1}{1} = \frac{z - 1}{- 1}$

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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

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x + 2 y - z - 5 = 0

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

π : 2 x - y + z - 4 = 0

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

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(- 1, 2, 0)

L_{1} : \frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}

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

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