A variable plane passes through a fixed point $\left(a,b,c\right)$ and cuts the coordinate axes at points $A,B,andC\dot{}$ Show that eh locus of the centre of the sphere $OABCis\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2.$

Solution: Let $\alpha ,\beta ,\gamma$ are coordinates of the locus at any point. So, at the centre of the sphere, equation can be given as:
${\left(x-\alpha \right)}^2+{\left(y-\beta \right)}^2+{\left(c-\gamma \right)}^2={\alpha }^2+{\beta }^2+{\gamma }^2$
(Here, ${\alpha }^2+{\beta }^2+{\gamma }^2={\left(radius\right)}^2$ as circle is passing through origin)
Solving the above equation, we get,
$\Rightarrow x^2+y^2+z^2-2\alpha x-2\beta y-2\gamma z=0$
At x-axis, it will be, $x^2=2\alpha x$ as y and z will be 0.
$x=2\alpha$
So, point at x-axis will be $\left(2\alpha ,0,0\right)$.
Similarly, at y-axis points will be$\left(0,2\beta ,0\right)$ and at z-axis, point will be $\left(0,0,2\gamma \right)$.
So, equation through plane at point (a,b,c) will be,
$\frac{a}{2\alpha }+\frac{b}{2\beta }+\frac{c}{2\gamma }=1$
$\Rightarrow \frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=2$
If, we replace $\left(\alpha ,\beta ,\gamma \right)$ by $\left(x,y,z\right)$, equation will be:
$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$