jmc

First note that$$\angle I_{1}A{I}_2=\angle I_{1}AX+\angle XA{I}_2=\frac{\angle BAX}{2}+\frac{\angle CAX}{2}=\frac{\angle A}{2}$$is a constant not depending on $X$, so by $[A{I}_1{I}_2]=\frac{1}{2}(A{I}_1)(A{I}_2)\sin \angle I_{1}A{I}_2$ it suffices to minimize $(A{I}_1)(A{I}_2)$. Let $a=BC$, $b=AC$, $c=AB$, and $\alpha =\angle AXB$. Remark that$$\angle A{I}_{1}B=180^{\circ }-(\angle I_{1}AB+\angle I_{1}BA)=180^{\circ }-\frac{1}{2}(180^{\circ }-\alpha )=90^{\circ }+\frac{\alpha }{2}.$$Applying the Law of Sines to $△AB{I}_1$ gives$$\frac{A{I}_1}{AB}=\frac{\sin \angle AB{I}_1}{\sin \angle A{I}_{1}B}\Rightarrow A{I}_1=\frac{c\sin \frac{B}{2}}{\cos \frac{\alpha }{2}}.$$Analogously one can derive $A{I}_2=\frac{b\sin \frac{C}{2}}{\sin \frac{\alpha }{2}}$, and so$$[A{I}_1{I}_2]=\frac{bc\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{2\cos \frac{\alpha }{2}\sin \frac{\alpha }{2}}=\frac{bc\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \alpha }\ge bc\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},$$with equality when $\alpha =90^{\circ }$, that is, when $X$ is the foot of the perpendicular from $A$ to $\overline{BC}$. In this case the desired area is $bc\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$. To make this feasible to compute, note that$$\sin \frac{A}{2}=\sqrt{\frac{1-\cos A}{2}}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}=\sqrt{\frac{(a-b+c)(a+b-c)}{4bc}}.$$Applying similar logic to $\sin \frac{B}{2}$ and $\sin \frac{C}{2}$ and simplifying yields a final answer of$$\begin{array}{rl}bc\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}& =bc\cdot \frac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\ & =\frac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{array}$$