Which point of the plane Ax+By+Cz+D=0 is closest to the…

Which point of the plane $Ax+By+Cz+D=0$ is closest to the origin?

Step 1: The perpendicular from $O$

The closest point is the foot of the perpendicular from $O$: $(x,y,z)=t\{A,B,C\}$. Substituting gives

\[ (A^2+B^2+C^2)\,t+D=0\ \Longrightarrow\ t=-\frac{D}{A^2+B^2+C^2}. \]

Step 2: Result

\[P_0\left(-\frac{AD}{A^2+B^2+C^2},\ -\frac{BD}{A^2+B^2+C^2},\ -\frac{CD}{A^2+B^2+C^2}\right)\]

\[ |OP_0|=|t|\,|\vec n|=\frac{|D|}{\sqrt{A^2+B^2+C^2}} . \]

By Pythagoras, any other point $X$ of the plane has $|OX|^2=|OP_0|^2+|P_0X|^2\gt |OP_0|^2$, so $P_0$ is the unique minimum. If $D=0$, then $P_0=O$.

Check: $A\cdot x_0+B\cdot y_0+C\cdot z_0+D=-D+D=0$. $\checkmark$