Question

A plane passes through (a,b, c) and cuts the axes in A,B,C, respectively, where
none of these points lie on the origin O. Show that the centre of the sphere OABC
satisfies the equation a
x +
b
y +
c
z = 2.

Answer 1

Let point $A\in Ox$ then $A(x_a;0;0)$ ,

$B\in Oy\implies B(0;x_b;0), \\ C\in Oz\implies C(0;0;z_c),O(0;0;0)$ .

Equation plane

$\frac{x}{x_a}+\frac{y}{y_b}+\frac{z}{z_c}=1$

Point $(a,b,c)$ in plane

$\frac{a}{x_a}+\frac{b}{y_b}+\frac{c}{z_c}=1$

Center of the sphere $D(l;m;n)$

$AD=BD=CD=OD=R$

$AD^2=(l-x_a)^2+(m-0)^2+(n-0)^2\\ BD^2=(l-0)^2+(m-y_b)^2+(n-0)^2\\ CD^2=(l-0)^2+(m-0)^2+(n-z_c)^2\\ OD^2=(l-0)^2+(m-0)^2+(n-0)^2$

$AD^2=OD^2\\ (l-x_a)^2+m^2+n^2=l^2+m^2+n^2\\ (l-x_a)^2=l^2\\ |l-x_a|=|l|\\ 1) l-x_a=l\\ l\in\empty, x_a=0\\ 2)l-x_a=-l\\ l=\frac{1}{2}x_a$

similarly

$BD^2=OD^2\\ l^2+(m-y_b)^2+n^2=l^2+m^2+n^2\\ (m-y_b)^2=m^2\\ m=\frac{1}{2}y_b\\ CD^2=OD^2\\ l^2+m^2+(n-z_c)^2=l^2+m^2+n^2\\ (n-z_c)^2=n^2\\ n=\frac{1}{2}z_c$

Then

$D(\frac{1}{2}x_a;\frac{1}{2}y_b;\frac{1}{2}z_c)=(l;m;n)\\ x_a=2l\\ y_b=2m\\ z_c=2n$

substitute into the equation

$\frac{a}{x_a}+\frac{b}{y_b}+\frac{c}{z_c}=1\\ \frac{a}{2l}+\frac{b}{2m}+\frac{c}{2n}=1\\ \frac{a}{l}+\frac{b}{m}+\frac{c}{n}=2\\$

where $(l,m,n)$  the coordinates centre of the sphere