The equation of line passing through $(1, - 2, 3)$ and having drs $(2, 3, 1)$ is

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

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

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

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none of these

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Solution

The correct option is A
$\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$

The point is $(1, - 2, 3)$

The drs are $(2, 3, 1)$

The equation is given as $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$

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Similar questions

(2, 3, 1)

x - 2 y - z + 5 = 0 and x + y + 3 z = 6

(- 4, 3, 1)

x + 2 y - z - 5 = 0

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

(- 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}

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

\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}

\frac{x - 3}{1} = \frac{y - 2}{- 4} = \frac{z}{5}

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