Problem:
Let $\,A,\,B,\,C,\,D\,$ be four distinct spheres in a space. Suppose the spheres $A$ and $B$ intersect along a circle which belongs to some plane $P$, the spheres $B$ and $C$ intersect along a circle which belongs to some plane $Q$, the spheres $C$ and $D$ intersect along a circle which belongs to some plane $S$, and the spheres $D$ and $A$ intersect along a circle which belongs to some plane $T$. Prove that the planes $\,P,\,Q,\,S,\,T\,$ are either parallel to the same line or have a commom point.
My Attempt:
Let us denote $\,\overrightarrow{AB}\,$ vector connecting centers of spheres $A$ and $B$, and $\overrightarrow{P}$ unit normal vector for plane $P$. Then, obviously, $\,\overrightarrow{AB}\perp{P}\iff \big\langle\overrightarrow{AB},\,\overrightarrow{P}\big\rangle=0\,$ Doing the same thing for the rest of planes and spheres, we get the system of equations:
\begin{align} \big\langle\overrightarrow{AB},\,\overrightarrow{P}\big\rangle = \big\langle\overrightarrow{BC},\,\overrightarrow{Q}\big\rangle = \big\langle\overrightarrow{CD},\,\overrightarrow{S}\big\rangle = \big\langle\overrightarrow{DA},\,\overrightarrow{T}\big\rangle &= 0 \end{align}
I do not know how to use these equations to prove that planes are either parallel to the same line, or have a commom point.
This is essentially the 3D equivalent of the Radical Center of 3 circles in 2D, and the proof works out in a similar way.
The plane through the intersection circle of two spheres $A, B$ is their radical plane, the locus of points with equal power $h_A = h_B$ with respect to the two spheres.
If the centers of spheres $A, B, C, D$ are all distinct, then all pairwise radical planes exist (not at infinity). This is obviously the case here, since the given spheres are known to intersect.
[1] Assuming that no two of the planes $P, Q, S, T$ are parallel, the intersection $P \cap Q$ will be a line. For all points $M$ on the line, $h_A = h_B$ since $M \in P$, and $h_B = h_C$ since $M \in Q$, so in the end $h_A = h_B = h_C$ for all points on $P \cap Q$.
[1.a] If line $P \cap Q$ is parallel to both planes $S$ and $T$, then that falls into the case where all 4 planes are parallel to a line $P, Q, S, T \parallel P \cap Q$.
[1.b] Otherwise, let WLOG $P \cap Q \not \parallel S$. Then the point of intersection $M = P \cap Q \cap S$ will also satisfy $h_C = h_D$. It follows that $h_A = h_B = h_C = h_D$, so $h_A = h_D$, therefore $M$ is also in the radical plane $T$ of circles $D, A$, and $M = P \cap Q \cap S \cap T$ is the common intersection point of the 4 radical planes.
[2] If two planes are parallel, let WLOG $P \parallel Q$. Since a radical plane is orthogonal to the line between the centers of the spheres, it follows that the centers $O_A, O_B, O_C$ are collinear.
[2.a] If the center $O_D$ of circle $D$ is on the same line $O_A O_B O_C$ then all 4 planes are parallel $P \parallel Q \parallel S \parallel T$.
[2.b] Otherwise, let $U$ be the plane defined by point $O_D$ and line $O_A O_B O_C$. Radical planes $S, T$ will be orthogonal to $U$, so all 4 planes will be parallel to the perpendicular on plane $U$.
Incidentally, this proves the proposition not just for the 4 planes considered here, but for all 6 pairwise radical planes between the 4 spheres.