Using Fact 5, we know that $I I_{A}$ intersects the circle $(A B C)$ at $M_{A}$, which is the center of $(I I_{A} B C X Y)$. Let $R$ be the radius of the latter circle. We have $R=\frac{1}{\sqrt{3}}$. We have $\angle A I M=\angle Y I I_{A}=\angle Y I X=\frac{\pi}{3}$. Also, $\angle I I_{A} M=\angle I M I_{A}$ by calculating the angles from the equilateral triangle. Using $90-60-30$ triangles, we have: $A I=\frac{1}{2} M I=\frac{1}{2} I I_{A}=R$, $A M=\frac{\sqrt{3}}{2} M I=\sqrt{3} R$, $M M_{A}^{2}=A M^{2}+A M_{A}^{2}=7 R^{2}$. Now, let $J$ and $N$ be the feet of the altitudes from $A$ and $B$ respectively on $M M_{A}$. Note that as $M$ is an arc midpoint of $B C, N$ is actually the midpoint of $B C$. $M_{A} J=\frac{A M_{A}^{2}}{M M_{A}}=\frac{4}{\sqrt{7}} R$, $M_{A} N=\frac{B M_{A}^{2}}{M M_{A}}=\frac{1}{\sqrt{7}} R$. Thus $J N=\frac{3}{\sqrt{7}} R$. Also, we have, $B N^{2}=M_{A} N \cdot M N=\frac{6}{7} R^{2}$. Now, $[A B C]=\frac{1}{2} J N \cdot B C=J N \cdot B N=\frac{3 \sqrt{6}}{7} R^{2}=\frac{\sqrt{6}}{7}$.