The vector equation of a plane which is at a distance of 9 units from the origin and which is normal to the
vector $2 ˆ i - ˆ j + 2 ˆ k$ is

\to r . (2 ˆ i - ˆ j + 2 ˆ k) = 27

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\to r . (2 ˆ i - ˆ j + 2 ˆ k) = 9

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\to r . (2 ˆ i - ˆ j + 2 ˆ k) = 3

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Solution

The correct option is A $\to r . (2 ˆ i - ˆ j + 2 ˆ k) = 27$
The vector equation of a plane is given in the form $\to r . ˆ n = d$
Here d = 9 and

$ˆ n = \frac{2 ˆ i - ˆ j + 2 ˆ k}{\sqrt{2^{2} + {(- 1)}^{2} + 2^{2}}} = \frac{2}{3} ˆ i - \frac{1}{3} ˆ j + \frac{2}{3} ˆ k$
So,
$- \to r . (\frac{2}{3} ˆ i - \frac{1}{3} ˆ j + \frac{2}{3} ˆ k) = 9$
$\to r . (2 ˆ i - ˆ j + 2 ˆ k) = 27 .$

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