Find the vector equation of a plane which is at a distance of $5$ units from the origin and its normal vector is $2^i - 3^j + 6^k$

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Solution

When equation of a plane is $a x + b y + c z + d = 0$ , its normal vector is $a^i + b^j + c^k$ and its distance from origin is $\frac{| d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$
Similarly, $a = 2, b = - 3, c = 6$ here.
Since $5 = \frac{| d |}{\sqrt{4 + 9 + 36}}$
$\Rightarrow | d | = 35$
The equation of the plane can thus be $2 x - 3 y + 6 z \pm 35 = 0$

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