The equation of the set of points P which is equidistant from the points (0, 2, 3) and (2, –2, 1) is:

x – 2y – z – 1 = 0

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x + 2y – z + 1 = 0

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x + 2y + z + 1 = 0

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x – 2y – z + 1 = 0

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Solution

The correct option is D
x – 2y – z + 1 = 0

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Similar questions

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

\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3}

(- 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}

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

x - 1 = 2 - y = z - 3

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

x - y + z + 2 = 0

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