Reduce the equation $x + 2 y - 3 z - 6 = 0$ of the plane in the normal form.

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Solution

$\begin{matrix} x + 2 y - 3 z = 6 \to r . (^i + 2^j - 3^k) = 6^r . \frac{(^i + 2^j - 3^k)}{\sqrt{14}} = \frac{6}{\sqrt{14}} \end{matrix}$
This is the required normal form.

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