The distance of the plane 6 x-2 y-3 z-12 from the origin is-A- 4B- 2C- 6D- 12-7

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Question

The distance of the plane 6x - 2y + 3z = 12 from the origin is:

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Solution

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A) 4
B) 5
C) 6
D) 7

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

x + 2 y + 3 z = 15

x - 2 y - 3 z - 3 \sqrt{14} = 0

C_{6} H_{14}

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