NDA 2025 Maths · Mathematics · Three Dimensional Geometry · easy

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

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The distance of the plane

2x-3y+6z=14

from the origin is

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Answer: (D) $2$

Distance

=\frac{|{-14}|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{14}{7}=2

. Writing the equation in the normal form

\frac27x-\frac37y+\frac67z=2

gives the same. The value

14

forgets to divide by

\sqrt{49}

7

is the denominator alone, and

\frac12

inverts the quotient.

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