The foot of the perpendicular drawn from the origin to a plane is $\left(8, -4, 3\right)$. Find the equation of the plane.
Let $O$ be the origin and $N \left(8, -4, 3\right)$ be the foot of perpendicular from $O$ to the given plane.
Let $P \left(x, y, z\right)$ be an arbitrary point on the plane.
Then,
direction ratios of $\overrightarrow{NP}$ are: $\;$ $\left(x - 8\right)$, $\left(y + 4\right)$, $\left(z - 3\right)$
direction ratios of $\overrightarrow{ON}$ are: $\;$ $8, \; -4, \; 3$
But $\overrightarrow{ON} \perp \overrightarrow{NP}$.
$\therefore$ $\;$ $\overrightarrow{ON} \cdot \overrightarrow{NP} = 0$
i.e. $\;$ $8 \left(x - 8\right) - 4 \left(y + 4\right) + 3 \left(z - 3\right) = 0$
$\therefore$ $\;$ the required equation of the plane is
$8x - 4y + 3z = 89$