Plane $SAB$ cuts $\Omega$ along a circle $(O)$ and cuts the circumference $\Omega \cap \omega$ again at $P.$ If $M,N$ denote the midpoints of $\overline{SA},\overline{SB},$ then $SMON$ is clearly cyclic with circumcircle $(K),$ the midcircle of $(O_1) \equiv \odot(SA_1B_1P)$ and $(O_2) \equiv \odot(SA_2B_2),$ whose center is then the midpoint of $\overline{O_1O_2}$ $\Longrightarrow$ $SO_1OO_2$ is a parallelogram with diagonal intersection $K$ $\Longrightarrow$ $SO_2 \parallel OO_1 \perp SP \parallel AB,$ i.e. $O_2$ is on S-altitude of $\triangle SAB,$ implying that $A_2B_2$ is antiparallel to $AB$ WRT $SA,SB$ $\Longrightarrow$ $SA \cdot SA_2=SB \cdot SB_2$ and similarly $SB \cdot SB_2=SC \cdot SC_2$ $\Longrightarrow$ $SA \cdot SA_2=SB \cdot SB_2=SC \cdot SC_2$ $\Longrightarrow$ $A,B,C,A_2,B_2,C_2,$ lie on a same sphere.

Let $O$ be the center of sphere $\omega$ and $O_1$ be the center of $\Omega$. Let $\mathcal{P}$ be the plane of the common circumference $\mathcal{C}$ of $\omega$ and $\Omega$. Let $O'$ be the center of $\mathcal{C}$ and note that $OO' \perp \mathcal{P}.$ Since $\mathcal{P}$ is parallel to the plane of $ABC$, we conclude that $O_1O' \perp \mathcal{P} \Longrightarrow O \in O'O_1$ so $OO_1$ is perpendicular to the plane of $ABC$. It follows that $OA=OB=OC$ so $A, B, C$ have an equal power $p$ in the sphere $\omega$. Evidently, we have $$p=AA_1 \cdot AS=SA \cdot SA_2=SB \cdot SB_2=SC \cdot SC_2,$$so we conclude that $A, B, C, A_2, B_2, C_2$ lie on a sphere.

Since $\omega ,\Omega$ intersect by plane, parallel to $(ABC),$ we easily conclude, that $A,B,C$ lie on a sphere, concentric with $\omega.$ Hence $|SA_{2}|\cdot |SA|=|AA_{1}|\cdot |AS|=|BB_{1}|\cdot |BS|=|SB_{2}|\cdot |SB|;$ this and analogous equalities yield desired.