It is equation (a) that is fulfilled.
Edit: Here is a simpler way to reach the result (the first draft is given as an appendix) wjth very few computations.
The variable plane passing through fixed point $F=(a,b,c)$ intersects the coordinate axes in $P=(p,0,0)$, $Q=(0,q,0)$ and $R=(0,0,r)$.
(we assume $p,q,r \ne 0$).
The equation of the plane $PQR$ is known to be :
$$\dfrac xp+\dfrac yq+\dfrac zr=1$$
As $F$ belongs to this plane, wh have :
$$\dfrac ap+\dfrac bq+\dfrac cr=1 \tag{1}$$
Let us now consider the box (parallelepiped) with vertices $O,P,Q,R,O',P',Q',R'$ where $O'=(p,q,r)$. The circumscribed sphere to this box, centered in
$$C=(x_0, y_0, z_0)=(\tfrac p2, \tfrac q2, \tfrac r2) \tag{2}$$
is clearly as well circumscribed to trirectangular tetrahedron $OPQR$, due to central symmetry with respect to $C$ ($P',Q',R'$ being the symmetrical points of $P,Q,R$ resp. with respect to $C$).
If we replace in (1) $(x,y,z)$ by $(x_0, y_0, z_0)$ defined by (2), we get relationship (a).
Appendix : first version : with the same notations as before.
Points $F,P,Q,R$ being coplanar, the following determinant is zero :
$$\begin{vmatrix}p& 0& 0 &a\\
0 &q& 0& b\\
0 &0& r& c\\
1 &1 &1&1\end{vmatrix}=0 \ \ \iff \ \ \dfrac ap+\dfrac bq+\dfrac cr=1 \tag{1}$$
(this is a classical coplanarity condition : see here).
Besides, the general equation of a sphere passing through the origin with center $(x_0, y_0, z_0)$ is :
$$x^2+y^2+z^2-2 x_0 x - 2 y_0 y -2 z_0 z=0$$
Let us express that this sphere passes through $P, Q, R$ resp.:
$$\begin{cases}p^2 - 2 x_0 p &=&0\\
q^2 - 2 y_0 q &=&0\\
r^2 - 2 z_0 r &=&0
\end{cases}$$
Therefore the coordinates of the center of the sphere are
$$(x_0, y_0, z_0)=(\tfrac p2, \tfrac q2, \tfrac r2)$$
which verify, taking account of (1) :
$$\dfrac{a}{2 x_0}+\dfrac{b}{2 y_0}+\dfrac{c}{2 z_0}=1$$
which is another way to write relationship (a).
Important remark : The center of the sphere has the remarkable geometric property to be the center of the box (= parallelepiped) defined by points $O,P,Q,R$ and $S=(p,q,r)$, [which is not in the base plane $PQR$ of "trirectangular tetrahedron" $OPQR$].
Best Answer
You have points $P(p,0,0),Q(0,q,0),R(0,0,r)$ on the axes and the origin $O$. Denote by $K$ the circumcenter of the triangle $PQR$. Since all the angles $\angle POQ,\angle QOR,\angle ROP$ are right, the sphere which passes through $P,Q,R,O$ has center $K$. To find $K$ easily, complete the $PQRO$ to a box, and $K$ would be then the midpoint of the great diagonal i.e. $K(p/2,q/2,r/2)$.
The plane has equation $k(x-a)+l(y-b)+m(z-c)=0$, (with variables $k,l,m$) written equivalently $\frac{x}{\frac{ka+lb+mc}{k}}+\frac{y}{\frac{ka+lb+mc}{l}}+\frac{z}{\frac{ka+lb+mc}{m}}=1$. This yields $p=\frac{ka+lb+mc}{k},q=\frac{ka+lb+mc}{l},r=\frac{ka+lb+mc}{m}$, and the coordinates of the center of the sphere are $x=\frac{ka+lb+mc}{2k},y=\frac{ka+lb+mc}{2l},\frac{ka+lb+mc}{2m}$.
Usually, if we use some classical geometry before we rush into many equations and calculations, we can solve some analytical geometry problems easier.