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

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

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None of these

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Solution

The correct option is A $\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 1}$
$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = k$
$\frac{x - 1}{a} = \frac{y + 2}{b} = \frac{z - 3}{c}$
$2 a + 3 b + c = 0$
This is possible when $a = 1, c = - 1, b = 1$
so, $2 (- 1) + 3 (1) + (- 1) = - 2 + 3 - 1 = 0$
$\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 1}$

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Similar questions

(- 4, 3, 1)

x + 2 y - z - 5 = 0

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

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

(- 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}

π : 2 x - y + z - 4 = 0

l

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

l

π

(2, 3, 1)

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

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