A plane passes through a fixed point $\left(a,b,c\right)\dot{}$ Show that the locus of the foot of the perpendicular to it from the origin is the sphere $x^2+y^2+z^2-ax-by-cz=0.$

Solution: Let M(x,y,z) is the foot of perpendicular and O(0,0,0) is the origin. We can draw figure from the given details.For figure, please refer to video.
From the figure, we can say that OM is perpendicular to PM.
It means
$\overrightarrow{OM}.\overrightarrow{MP}=0\to Eq\left(1\right)$
$\overrightarrow{OM}=x\hat{i}+y\hat{j}+z\hat{k}$
$\overrightarrow{MP}=\left(a-x\right)\hat{i}+\left(b-y\right)\hat{j}+\left(c-z\right)\hat{k}$
Now, putting these value in Eq(1)
$\left(x\hat{i}+y\hat{j}+z\hat{k}\right)\left(\left(a-x\right)\hat{i}+\left(b-y\right)\hat{j}+\left(c-z\right)\hat{k}\right)=0$
$\Rightarrow x\left(a-x\right)+y\left(b-y\right)+z\left(c-z\right)=0$
$x^2+y^2+z^2-ax-by-cz=0$ that is the required equation.