If the foot of perpendicular drawn from the origin to the plane is \((3,2,1)\), then the equation of plane is

(B)
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
\(\begin{aligned}
&(x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14} \\
\therefore & 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0
\end{aligned}\)