Define an ordered triple $ (A, B, C)$ of sets to be minimally intersecting if $ |A \cap B| \equal{} |B \cap C| \equal{} |C \cap A| \equal{} 1$ and $ A \cap B \cap C \equal{} \emptyset$. For example, $ (\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $ N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $ \{1,2,3,4,5,6,7\}$. Find the remainder when $ N$ is divided by $ 1000$.

[b]Note[/b]: $ |S|$ represents the number of elements in the set $ S$.
1. **Choosing the elements for intersections:**
   - We need to choose three distinct elements from the set \(\{1, 2, 3, 4, 5, 6, 7\}\) to be the elements of \(A \cap B\), \(B \cap C\), and \(C \cap A\). 
   - The number of ways to choose 3 distinct elements from 7 is given by the combination formula \(\binom{7}{3}\):
     \[
     \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
     \]

2. **Arranging the chosen elements:**
   - Once we have chosen the 3 elements, we need to assign them to \(A \cap B\), \(B \cap C\), and \(C \cap A\). There are \(3!\) (3 factorial) ways to arrange these 3 elements:
     \[
     3! = 6
     \]

3. **Choosing the remaining elements:**
   - There are 4 remaining elements in the set \(\{1, 2, 3, 4, 5, 6, 7\}\) after choosing the 3 elements for intersections.
   - Each of these 4 elements can either be in set \(A\), set \(B\), set \(C\), or in none of the sets. Thus, each element has 4 choices.
   - The total number of ways to distribute these 4 elements is:
     \[
     4^4 = 256
     \]

4. **Calculating the total number of minimally intersecting triples:**
   - Multiply the number of ways to choose and arrange the 3 elements by the number of ways to distribute the remaining 4 elements:
     \[
     35 \times 6 \times 256 = 53760
     \]

5. **Finding the remainder when \(N\) is divided by 1000:**
   - We need to find \(53760 \mod 1000\):
     \[
     53760 \div 1000 = 53 \text{ remainder } 760
     \]
   - Therefore, the remainder is 760.

The final answer is \(\boxed{760}\).