The equation of the line passing through (1, 2, 3) and parallel to the planes x - y + 2z = 5 and 3x + y + z = 6, is
[DSSE 1986]

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Solution

The correct option is A

$\frac{x - 1}{l} = \frac{y - 2}{m} = \frac{z - 3}{n}$
or l - m + 2n = 0 and 3l + m + n = 0
$∴ \frac{x - 1}{- 3} = \frac{y - 2}{5} = \frac{z - 3}{4}$ .

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Similar questions

(1, - 2, 3)

x - y + 2 z - 5 = 0

3 x + y + z - 6 = 0

(1, 2, 3)

\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2 z - 6}{3}

Equation of line passing through the point (2, 3, 1) and parallel to the line of intersection of the plane x – 2y – z + 5 = 0 and x + y + 3z = 6 is

2 x - 3 y + 6 z = 11

x - y + 2 z = 5

3 x + y + z = 6

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