We show that the point $J$, usually denoted as $H$, the orthocenter of triangle $ABC$, has the described property. For this, let $a$ be the line through $A$ parallel to $BC$, and the lines $b$ and $c$ are defined analogously. No two of the three lines $a, b$, and $c$ are parallel, and therefore, the intersection point $A'$ of $b$ with $c$ and the two analogously defined points $B'$ and $C'$ exist.
The quadrilateral $A' C J B$ has right angles at $B$ and $C$ and is therefore a cyclic quadrilateral. Since the reflection at the midpoint of the segment $\overline{BC}$ maps point $A$ to $A'$ and swaps $B$ with $C$, it transforms $\omega$ into the circumcircle of the just found cyclic quadrilateral. Consequently, $A_2$ lies on this circle, and $\Lambda' A_2$ is parallel to $AX$.
Let the centroid of triangle $ABC$ be $S$. The central dilation $\sigma$ with center $S$ and factor $-2$ maps $A$ to $A'$; the image of $X$ is denoted by $X'$. Then, the lines $AX$ and $A'X''$ are parallel, and therefore, $X'$ lies on the line $A'A_2$. Similar arguments can be made by replacing $A$ with $B$ and $C$, and we learn in total: The three lines $A'A_2, B'B_2$, and $C''C_2$ intersect at $X'$.
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If $X' = J$, the points $A_2, B_2$, and $C_2$ coincide with $J$, and the claim is trivial. From now on, let $X' \neq J$.
Since $\varangle A' C J = 90^\circ$, the segment $\overline{A' J}$ is, by Thales' theorem, a diameter of the circumcircle of the quadrilateral $A' C J B$. Again, by Thales' theorem, $\varangle J A_2 A' = 90^\circ$ and therefore also $\varangle J A_2 X' = 90^\circ$. Thus, the point $A_2$ lies, once again by Thales' theorem, on the circle with diameter $J X^{-1}$. For similar reasons, the points $B_2$ and $C_2$ also lie on this circle, and in particular, we have now found a circle on which all four points $A_2, B_2, C_2$, and $J$ lie.