Let $\Omega_1(O_1,r_1)$ and $\Omega_2(O_2,r_2)$ denote the spheres through $A,B,C$ internally tangent to $\mathcal{S}(O,R)$ at $U,V.$ Since $AB \perp OA$ and $AC \perp OA,$ then the plane $ABC$ is perpendicular to $OA$ $\Longrightarrow$ $O_1O_2 \parallel OA$ $\Longrightarrow$ $O_1,O_2,O,A$ lie on a same plane $\delta,$ which obviously contains $U,V,$ because of $U \in OO_1$ and $V \in OO_2.$ This plane $\delta$ cuts then $\Omega_1,\Omega_2,\mathcal{S}$ into great circles $(O_1,r_1),(O_2,r_1),(O,R)$ such that $A \equiv (O_1) \cap (O_2)$ and $(O)$ touches $(O_1),(O_2)$ internally at $U,V.$ WLOG assume that $r_1 > r_2.$ Then $OO_2-OO_1=R-r_2-(R-r_1)=r_1-r_2=AO_1-AO_2$ $\Longrightarrow$ $O$ and $A$ lie on different branches of the hyperbola $\mathcal{H}$ with foci $O_1,O_2,$ major axis length $(r_1-r_2)$ and conjugate axis the perpedicular bisector $\ell$ of $\overline{O_1O_2}.$ Since $\mathcal{H}$ is symmetric about $\ell,$ then $OA \parallel O_1O_2$ implies that $O$ is the reflection of $A$ about $\ell,$ i.e. $AOO_1O_2$ is an isosceles trapezoid with legs $OO_1=AO_2$ $\Longrightarrow$ $R-r_1=r_2.$