The equation of the line passing through the point (1, 1, –1) and perpendicular to the plane x – 2y – 3z = 7 is

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

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

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

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

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Solution

The correct option is C $\frac{x - 1}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{- 3}$
Line is parallel to the normal of the plane x – 2y – 3z = 7
$∴$ Equation of the line through (1, 1, –1) is
$\frac{x - 1}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{- 3}$

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