Japan competition problems, 2015

$$\frac{288}{35}$$
For a polygon $P$, let us denote by $S(P)$ its area.
Let $I$ be the incenter of the circle $\Gamma$. Since $AI$ is the bisector of the angle $\angle CAB$, $I$ lies on the line segment $AA'$. Similarly, $I$ lies on the line segments $BB'$ and $CC'$. By using the theorem on angles at a point on the circumference of a circle subtended by arcs of the circle, we have
$$\angle C'B'A = \angle C'CA = \angle C'CB = \angle C'B'B = \angle C'B'I.$$
Similarly, we obtain $\angle B'C'A = \angle B'C'I$. We then see that the triangles $AB'C'$ and $IB'C'$ are congruent. For the same reason, we have that the triangles $A'BC'$ and $A'IC'$ are congruent and so are the triangles $A'B'C$ and $A'B'I$. From these facts we obtain that $S(IB'C') = 2$, $S(IC'A') = 3$ and $S(IA'B') = 4$, from which it follows that
$$\begin{aligned} AI : IA' &= S(AC'IB') : S(A'C'IB') \\ &= (2+2) : (3+4) = 4 : 7. \end{aligned}$$
Similarly, we obtain
$$BI : IB' = (3+3) : (4+2) = 1 : 1$$
$$CI : IC' = (4+4) : (2+3) = 8 : 5.$$
Utilizing these, we obtain
$$S(BIC) = S(B'IC') \cdot \frac{BI}{IB'} \cdot \frac{CI}{IC'} = 2 \cdot \frac{1}{1} \cdot \frac{8}{5} = \frac{16}{5}$$
$$S(CIA) = S(C'IA') \cdot \frac{CI}{IC'} \cdot \frac{AI}{IA'} = 3 \cdot \frac{8}{5} \cdot \frac{4}{7} = \frac{96}{35}$$
$$S(AIB) = S(A'IB') \cdot \frac{AI}{IA'} \cdot \frac{BI}{IB'} = 4 \cdot \frac{4}{7} \cdot \frac{1}{1} = \frac{16}{7}$$
Therefore, the answer we seek is $\frac{16}{5} + \frac{96}{35} + \frac{16}{7} = \frac{288}{35}$.