If the foot of perpendicular drawn from the origin to the plane is $(3,2,1)$, then the equation of plane is
If the foot of perpendicular drawn from the origin to the plane is $(3,2,1)$, then the equation of plane is
$3x+2y-z=12$
$3x+2y-z=14$
$3x+2y+z=14$
$3x-2y-z=12$
Solution
Let $P(3,2,1)$ be the foot of the perpendicular from the origin to the plane.
$\begin{aligned}
\overline{\mathrm{OP}} &=\overline{\mathrm{n}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and } \\
\mathrm{P} &=|3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3^{2}+2^{2}+1^{2}}=\sqrt{14} \\
\hat{\mathrm{n}} &=\frac{3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}
\end{aligned}$
The vector equation of the plane is
$\begin{aligned}
&(x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14} \\
\therefore & 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0
\end{aligned}$