In triangle $ABC$, $\sin A \sin B \sin C = \frac{1}{1000}$ and $AB \cdot BC \cdot CA = 1000$.  What is the area of triangle $ABC$?

[i]Proposed by Evan Chen[/i]
1. **Given Information:**
   - \(\sin A \sin B \sin C = \frac{1}{1000}\)
   - \(AB \cdot BC \cdot CA = 1000\)

2. **Using the Extended Law of Sines:**
   The extended law of sines states that for any triangle \(ABC\) with circumradius \(R\),
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   where \(a = BC\), \(b = CA\), and \(c = AB\).

3. **Relating the Given Equations:**
   We need to find the circumradius \(R\). We start by dividing the product of the side lengths by the product of the sines of the angles:
   \[
   \frac{AB \cdot BC \cdot CA}{\sin A \cdot \sin B \cdot \sin C} = \frac{1000}{\frac{1}{1000}} = 1000^2 = 10^6
   \]

4. **Using the Extended Law of Sines:**
   From the extended law of sines, we know:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   Therefore,
   \[
   AB \cdot BC \cdot CA = (2R \sin A) \cdot (2R \sin B) \cdot (2R \sin C) = (2R)^3 \sin A \sin B \sin C
   \]
   Substituting the given values:
   \[
   1000 = (2R)^3 \cdot \frac{1}{1000}
   \]
   Simplifying, we get:
   \[
   1000^2 = (2R)^3
   \]
   \[
   10^6 = 8R^3
   \]
   \[
   R^3 = \frac{10^6}{8} = 125000
   \]
   \[
   R = \sqrt[3]{125000} = 50
   \]

5. **Finding the Area of Triangle \(ABC\):**
   The area of triangle \(ABC\) can be found using the formula:
   \[
   [ABC] = \frac{abc}{4R}
   \]
   Substituting the known values:
   \[
   [ABC] = \frac{1000}{4 \cdot 50} = \frac{1000}{200} = 5
   \]

The final answer is \(\boxed{5}\).