Find all three-digit numbers $ abc$ (with $ a \neq 0$) such that $ a^{2}+b^{2}+c^{2}$ is a divisor of 26.
To solve the problem, we need to find all three-digit numbers \( abc \) such that \( a^2 + b^2 + c^2 \) is a divisor of 26. The divisors of 26 are 1, 2, 13, and 26. We will consider each case separately.

1. **Case 1: \( a^2 + b^2 + c^2 = 1 \)**
   - Since \( a \neq 0 \) and \( a, b, c \) are digits (0-9), the only possibility is \( a = 1 \) and \( b = c = 0 \).
   - Therefore, the valid number is \( 100 \).

2. **Case 2: \( a^2 + b^2 + c^2 = 2 \)**
   - Since \( a \neq 0 \), \( a \) must be 1 (as \( 2 = 1^2 + 1 \)).
   - The remaining sum \( b^2 + c^2 = 1 \), which implies \( b \) and \( c \) can be 1 and 0 in any order.
   - Therefore, the valid numbers are \( 110 \) and \( 101 \).

3. **Case 3: \( a^2 + b^2 + c^2 = 13 \)**
   - We need to find digits \( a, b, c \) such that their squares sum to 13.
   - Since \( a \neq 0 \), we consider \( a \) values from 1 to 3 (as \( 4^2 = 16 > 13 \)).
     - If \( a = 3 \), then \( b^2 + c^2 = 13 - 9 = 4 \). The pairs \((b, c)\) that satisfy this are \((2, 0)\) and \((0, 2)\).
     - Therefore, the valid numbers are \( 320 \), \( 302 \), \( 230 \), and \( 203 \).

4. **Case 4: \( a^2 + b^2 + c^2 = 26 \)**
   - We need to find digits \( a, b, c \) such that their squares sum to 26.
   - Since \( a \neq 0 \), we consider \( a \) values from 1 to 5 (as \( 6^2 = 36 > 26 \)).
     - If \( a = 5 \), then \( b^2 + c^2 = 26 - 25 = 1 \). The pairs \((b, c)\) that satisfy this are \((1, 0)\) and \((0, 1)\).
     - Therefore, the valid numbers are \( 510 \), \( 501 \), \( 150 \), and \( 105 \).
     - If \( a = 4 \), then \( b^2 + c^2 = 26 - 16 = 10 \). There are no pairs of digits \((b, c)\) that satisfy this.
     - If \( a = 3 \), then \( b^2 + c^2 = 26 - 9 = 17 \). There are no pairs of digits \((b, c)\) that satisfy this.
     - If \( a = 2 \), then \( b^2 + c^2 = 26 - 4 = 22 \). There are no pairs of digits \((b, c)\) that satisfy this.
     - If \( a = 1 \), then \( b^2 + c^2 = 26 - 1 = 25 \). The pairs \((b, c)\) that satisfy this are \((5, 0)\) and \((0, 5)\).
     - Therefore, the valid numbers are \( 150 \) and \( 105 \).

Combining all the valid numbers from each case, we get the following three-digit numbers: \( 100, 110, 101, 320, 302, 230, 203, 510, 501, 150, 105 \).

The final answer is \( \boxed{100, 110, 101, 320, 302, 230, 203, 510, 501, 150, 105} \).