We use a spatial inversion (just like in 2D) which has one of the 3 pts of tangency as pole. It turns the 2 circles which pass through the pole into lines, which are both tangent to the circle which is the image of the 3'rd circle, so the images of the 3 circles are coplanar (a circle and 2 lines tangent to it), which means that the points were originally on a sphere (which passes through the point chosen as pole). It's a nice problem!

nice solution Grobber but more elementary one is : let $w_1,w_2,w_3$ be three circles ,pairwisely tangent to each others at $P_3,P_2,P_1,$ respectively. Lemma: let $X,Y$ be two distinct points on the circle $Q$ in $3-D.$ let $S$ be an arbitrary sphere passing through $X,Y.$ then there exist a set of points common in $Q,S.$ proof: just let $O$ be the circle with diameter of $S$ which passes through $X,Y$;so lemma proved. case 1: not all of the circles tangent externaly, so there exist two circles WLOG called $w_1,w_2$ which the projection of them on the plane of $w_3,$ is completely into the circle $w_3.$ as we know not all of them are coplanar, so $P_1,P_2,P_3$ make a triangle. let $W$ be the circumcircle of this triangle.rotate $W$ from one of it's diameters to build a sphere>>>Lemma case 2: all of the circles are externaly tangent to each others, which means projection of no two circles into the plane of the third one,is completely in it. again because not all of the circles are coplanar, so $P_1,P_2,P_3$ make a triangle.again let $W$ be the circumcircle of this triangle.rotate $W$ from one of it's diameters to build a sphere>>>Lemma. clarify of an assumption : because not all of the circles are coplanar, so $P_1,P_2,P_3$ make a triangle. there's a case when the centers of $w_1,w_2,w_3$ are collinear,and so $P_1,P_2,P_3$ are collinear. but circles are not coplanar,so the assumption is correct.

grobber,can you tell me what spatial inversion is? BTW is inversion also applicable in three dimensional space?