Answer: 126

Solution:

Since $AI_1$ and $AI_2$ bisect angles $BAX$ and $XAC,$ $\angle I_1AI_2 = \frac{1}{2}\angle BAX$ $+ \frac{1}{2}\angle XAC = \frac{A}{2},$ a constant. Let $\alpha = \angle AXB.$ The incenter angle formula gives $\angle AI_1B = 90^\circ + \frac{\alpha}{2},$ so the law of sines in $\triangle ABI_1$ yields $AI_1 = \frac{AB \sin\frac{B}{2}}{\cos\frac{\alpha}{2}},$ and similarly, since $\angle AXC = 180^\circ - \alpha,$ $AI_2 = \frac{AC \sin\frac{C}{2}}{\sin\frac{\alpha}{2}}.$

Therefore $$\begin{aligned} &[\triangle AI_1I_2] = \frac{1}{2}\,AI_1 \cdot AI_2 \sin\frac{A}{2} \\ &= \frac{AB \cdot AC \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\alpha}, \end{aligned}$$ which is minimized when $\alpha = 90^\circ,$ that is, when $X$ is the foot of the altitude from $A.$

With $a = 32,$ $b = 34,$ $c = 30,$ and $s = 48,$ the half-angle formulas give $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ $= \frac{(s-a)(s-b)(s-c)}{abc},$ so the minimum area is $$\begin{aligned} &bc \cdot \frac{(s-a)(s-b)(s-c)}{abc} \\ &= \frac{16 \cdot 14 \cdot 18}{32} = 126. \end{aligned}$$