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Let $6 x-3 y+2 z-6=0$ be the given plane. If $a, b, c$ are the intercepts made by the plane on $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$-axes respectively; $1, m, n$ are the direction cosines of a normal drawn to the plane and $\mathrm{p}$ is the perpendicular distance from the origin to the plane, then $|a l+b m+c n|=$
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The correct answer is:
$3 \mathrm{p}$
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