The distance of the plane 6x - 3y + 2z - 14 = 0 from the origin is:

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Solution

Q.
The distance of the plane $6 x - 3 y + 2 z - 14 = 0$ from the origin is:

1 2

2 1

3 14

4 8

Ans.
(1)

Ans. 2

Explanation:

Distance of plane from origin

$= \frac{d}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{14}{\sqrt{6^{2} + 3^{2} + 2^{2}}}$

$= \frac{14}{\sqrt{49}} = \frac{14}{7} = 2$

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