Denote by $J_a$, $J_b$, and $J_c$ the centers of the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. We prove that $J_a$, $J_b$, and $J_c$ are collinear.

Let $L_a$, $L_b$, and $L_c$ be the points diametrically opposite to $I$ on the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. Since $J_a J_b$, $J_b J_c$, and $J_c J_a$ are midlines in triangles $IL_a L_b$, $IL_b L_c$, and $IL_c L_a$, respectively, it suffices to show that $L_a$, $L_b$, and $L_c$ are collinear.

Since $IL_a$ is a diameter in the circumcircle of triangle $IEI_a$, we have $\angle II_a L_a = 90^\circ$. Hence $I_a L_a \perp II_a$. But $II_a \perp I_b I_c$, so $I_a L_a \parallel I_b I_c$, implying
$$\angle L_a I_a B = \angle I_a I_c I_b, \quad \text{and} \quad \angle L_a I_a C = 180^\circ - \angle I_a I_b I_c. \qquad (1)$$
Let $J$ be the midpoint of segment $IE$. Since $JJ_a$ is a midline in triangle $IEL_a$, it follows that $JJ_a \parallel EL_a$. But $JJ_a$ is the perpendicular bisector of $IE$, so $JJ_a \perp IE$, and since $BE \perp IE$, it follows that $JJ_a \parallel BE$. From $BE \parallel JJ_a$ and $EL_a \parallel JJ_a$, we deduce that $L_a \in BE$, so $L_a \in BC$. Similarly, we deduce that $L_b \in CA$ and $L_c \in AB$.

We have $\frac{BL_a}{CL_a} = \frac{S_{BI_a L_a}}{S_{CI_a L_a}} = \frac{BI_a \cdot L_a I_a \cdot \sin(\angle L_a I_a B)}{CI_a \cdot L_a I_a \cdot \sin(\angle L_a I_a C)} \stackrel{(1)}{=} \frac{BI_a \cdot \sin(\angle I_a I_c I_b)}{CI_a \cdot \sin(\angle I_a I_b I_c)}$.
Similarly, $\frac{CL_b}{AL_b} = \frac{CI_b \cdot \sin(\angle I_b I_a I_c)}{AI_b \cdot \sin(\angle I_b I_c I_a)}$, $\frac{AL_c}{BL_c} = \frac{AI_c \cdot \sin(\angle I_c I_b I_a)}{BI_c \cdot \sin(\angle I_c I_a I_b)}$.

Therefore, $\frac{BL_a}{CL_a} \cdot \frac{CL_b}{AL_b} \cdot \frac{AL_c}{BL_c} = 1$, and by the converse of Menelaus' theorem, $L_a$, $L_b$, and $L_c$ are collinear, hence $J_a$, $J_b$, and $J_c$ are collinear.
It follows that the radical axes of any two of the three circles are parallel or coincide. Since $I$ lies on all three circles, we deduce that these circles have the same radical axis. But $I \notin J_aJ_b$, so the three circles are not tangent, and therefore their radical axis contains a point different from $I$ lying on all three circles.