In an equilateral triangle of side length 6 , pegs are placed at the vertices and also evenly along each side at a distance of 1 from each other. Four distinct pegs are chosen from the 15 interior pegs on the sides (that is, the chosen ones are not vertices of the triangle) and each peg is joined to the respective opposite vertex by a line segment. If $N$ denotes the number of ways we can choose the pegs such that the drawn line segments divide the interior of the triangle into exactly nine regions, find the sum of the squares of the digits of $N$.
1. **Identify the total number of pegs:**
   - The equilateral triangle has side length 6.
   - Pegs are placed at the vertices and evenly along each side at a distance of 1 from each other.
   - Each side has 7 pegs (including the vertices).
   - Total pegs on the sides = \(3 \times 7 = 21\).
   - Excluding the vertices, the number of interior pegs on the sides = \(21 - 3 = 18\).

2. **Choose 4 distinct pegs from the 15 interior pegs:**
   - We need to choose 4 pegs from the 15 interior pegs (excluding the vertices).
   - The number of ways to choose 4 pegs from 15 is given by the binomial coefficient:
     \[
     \binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4! \cdot 11!} = 1365
     \]

3. **Consider the two cases for the division into 9 regions:**
   - **Case 1: Three cevians are concurrent and another one passes through one of the points.**
     - By Ceva's theorem, for some positive integers \(a, b, c\) in between 1 to 5:
       \[
       \frac{a}{6-a} \cdot \frac{b}{6-b} \cdot \frac{c}{6-c} = 1 \implies abc = (6-a)(6-b)(6-c)
       \]
     - From \(a, b, c \in \{1, 2, 3, 4, 5\}\), there are 13 such possible triplets.
     - The fourth cevian can be joined to any of the other 12 interior points.
     - Total possible ways for this case:
       \[
       13 \cdot \binom{12}{1} = 13 \cdot 12 = 156
       \]

   - **Case 2: Choose any two sides and any two interior points from each.**
     - Choose 2 sides from 3:
       \[
       \binom{3}{2} = 3
       \]
     - Choose 2 interior points from 5 on each chosen side:
       \[
       \binom{5}{2} = 10
       \]
     - Total possible ways for this case:
       \[
       3 \cdot 10 \cdot 10 = 300
       \]

4. **Calculate the total number of ways:**
   - Sum the number of ways from both cases:
     \[
     300 + 156 = 456
     \]

5. **Find the sum of the squares of the digits of \(N\):**
   - The digits of 456 are 4, 5, and 6.
   - Sum of the squares of the digits:
     \[
     4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77
     \]

The final answer is \(\boxed{77}\)