Centre of the sphere

Question

A variable plane passes through a fixed point $(a,b,c)$ and cuts the coordinate axes at $P,Q,R$. Then the coordinates $(x,y,z)$ of the centre of the sphere passing through $P,Q,R$ and the origin satisfy which of the following equation?
(A) $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
(B) $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3$
(C) $ax+by+cz=1$
(D) $ax+by+cz=a^2+b^2+c^2$
I haven't been able to make much progress in this question. We can take the equation of the plane to be $a_1(x-a)+b_1(y-b)+c_1(z-c)=0$. Then we can find $P,Q,R$. Then I am not sure what to do next.
P.S.- I don't know the general equation of a sphere.

Answer 1

To follow my comment:

Your plane is $Ax+By+Cz=Aa+Bb+Cc=D$ - to check, see that the given point lies on the plane, and that this is a linear form.

It cuts the co-ordinate axes at $(\frac DA,0,0); (0,\frac DB,0); (0,0,\frac DC)$

The centre of the circle at which the sphere cuts the plane $z=0$ is half way along the line between $(\frac DA,0,0)$ and $(0,\frac DB,0)$ i.e. at $(\frac {D}{2A},\frac {D}{2B},0)$.

The perpendicular to the plane at this point is (easy because it is a co-ordinate plane) $x=\frac {D}{2A}; y=\frac {D}{2B}$

You should be able to use this method to work out the co-ordinates of the centre of the sphere. Then check which equation the centre will satisfy.