If `Delta=|(1,3cos theta,1),(sin theta,1,3cos theta),(1,sin theta,1)|,` then the `1/1000` [|maximum value of `Delta-` minimum value of `Delta|^3`] is equal to.

Answer

To solve the problem, we need to calculate the determinant \(\Delta\) and then find the maximum and minimum values of \(\Delta\). Finally, we will compute \(\frac{1}{1000} \left| \text{maximum value of } \Delta - \text{minimum value of } \Delta \right|^3\).

### Step 1: Calculate the Determinant \(\Delta\)

Given:
\[ \Delta = \begin{vmatrix} 1 & 3\cos\theta & 1 \\ \sin\theta & 1 & 3\cos\theta \\ 1 & \sin\theta & 1 \end{vmatrix} \]

We can expand this determinant using the first row:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & 3\cos\theta \\ \sin\theta & 1 \end{vmatrix} - 3\cos\theta \cdot \begin{vmatrix} \sin\theta & 3\cos\theta \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} \sin\theta & 1 \\ 1 & \sin\theta \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 1 & 3\cos\theta \\ \sin\theta & 1 \end{vmatrix} = 1 \cdot 1 - 3\cos\theta \cdot \sin\theta = 1 - 3\sin\theta\cos\theta\)

2. \(\begin{vmatrix} \sin\theta & 3\cos\theta \\ 1 & 1 \end{vmatrix} = \sin\theta \cdot 1 - 3\cos\theta \cdot 1 = \sin\theta - 3\cos\theta\)

3. \(\begin{vmatrix} \sin\theta & 1 \\ 1 & \sin\theta \end{vmatrix} = \sin^2\theta - 1\)

Putting these back into the determinant:
\[ \Delta = 1 - 3\sin\theta\cos\theta - 3\cos\theta(\sin\theta - 3\cos\theta) + (\sin^2\theta - 1) \]
\[ = 1 - 3\sin\theta\cos\theta - 3\sin\theta\cos\theta + 9\cos^2\theta + \sin^2\theta - 1 \]
\[ = \sin^2\theta + 9\cos^2\theta - 6\sin\theta\cos\theta \]

### Step 2: Simplify \(\Delta\)

Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\[ \Delta = 1 + 8\cos^2\theta - 6\sin\theta\cos\theta \]

### Step 3: Find Maximum and Minimum Values of \(\Delta\)

Let \(x = \cos 2\theta\) and \(y = \sin 2\theta\). We can express \(\Delta\) as:
\[ \Delta = 5 + 4\cos 2\theta - 3\sin 2\theta \]

To find the maximum and minimum values, we can use the formula for the maximum and minimum of a linear combination of sine and cosine:
\[ R = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = 5 \]

Thus,
\[ -5 \leq 4\cos 2\theta - 3\sin 2\theta \leq 5 \]

Adding 5 to all parts:
\[ 0 \leq 5 + 4\cos 2\theta - 3\sin 2\theta \leq 10 \]

So, the maximum value of \(\Delta\) is 10 and the minimum value is 0.

### Step 4: Calculate \(\frac{1}{1000} \left| \text{maximum value of } \Delta - \text{minimum value of } \Delta \right|^3\)

Now we compute:
\[ \frac{1}{1000} \left| 10 - 0 \right|^3 = \frac{1}{1000} \cdot 10^3 = \frac{1000}{1000} = 1 \]

### Final Answer
The final answer is:
\[ \boxed{1} \]