A plane passes through a fixed point (a, b, c). The locus of the foot of the perpendicular to it from the origin is a sphere of radius

$\text{ Let the foot of the perpendicular from the origin on }$$\text{ the given plane be }P(\alpha ,\beta ,\gamma )$

$\text{ The plane passes through }A(a,b,c)$

$\begin{array}{l}\therefore AP\perp OP\\ \Rightarrow \overrightarrow{AP}\cdot \overrightarrow{OP}=0\\ \Rightarrow \left[\right(\alpha -a)\widehat{i}+(\beta -b)\widehat{j}+(\gamma -c\left)\widehat{k}\right]\cdot (\alpha \widehat{i}+\beta \widehat{j}+\gamma \widehat{k})=0\\ \alpha (\alpha -a)+\beta (\beta -b)+\gamma (\gamma -c)=0\end{array}$

$\text{ Hence, the locus of }(\alpha ,\beta ,\gamma )\text{ is }$

$\begin{array}{r}x(x-a)+y(y-b)+z(z-c)=0\\ or x^2+y^2+z^2-ax-by-cz=0\end{array}$

$\text{ which is a sphere of radius }\frac{1}{2}\sqrt{a^2+b^2+c^2}\text{. }$