Find all non-negative integers $a, b, c$ such that the roots of equations: $\begin{cases}x^2 - 2ax + b = 0 \\
x^2- 2bx + c = 0 \\
x^2 - 2cx + a = 0 \end{cases}$ are non-negative integers.
We need to find all non-negative integers \(a, b, c\) such that the roots of the following equations are non-negative integers:
\[
\begin{cases}
x^2 - 2ax + b = 0 \\
x^2 - 2bx + c = 0 \\
x^2 - 2cx + a = 0 
\end{cases}
\]

1. **Case 1: \(a, b, c > 0\)**

   For the roots of the quadratic equations to be non-negative integers, the discriminants of these quadratics must be perfect squares. Let's analyze each equation:

   - For \(x^2 - 2ax + b = 0\), the discriminant is:
     \[
     \Delta_1 = (2a)^2 - 4b = 4a^2 - 4b
     \]
     For \(\Delta_1\) to be a perfect square, \(4a^2 - 4b\) must be a perfect square. Let \(4a^2 - 4b = k^2\) for some integer \(k\). This implies:
     \[
     4(a^2 - b) = k^2 \implies a^2 - b = \left(\frac{k}{2}\right)^2
     \]
     Since \(a, b\) are integers, \(\frac{k}{2}\) must be an integer, so \(k\) must be even. Let \(k = 2m\), then:
     \[
     a^2 - b = m^2 \implies b = a^2 - m^2
     \]

   - For \(x^2 - 2bx + c = 0\), the discriminant is:
     \[
     \Delta_2 = (2b)^2 - 4c = 4b^2 - 4c
     \]
     Similarly, for \(\Delta_2\) to be a perfect square, \(4b^2 - 4c\) must be a perfect square. Let \(4b^2 - 4c = n^2\) for some integer \(n\). This implies:
     \[
     4(b^2 - c) = n^2 \implies b^2 - c = \left(\frac{n}{2}\right)^2
     \]
     Since \(b, c\) are integers, \(\frac{n}{2}\) must be an integer, so \(n\) must be even. Let \(n = 2p\), then:
     \[
     b^2 - c = p^2 \implies c = b^2 - p^2
     \]

   - For \(x^2 - 2cx + a = 0\), the discriminant is:
     \[
     \Delta_3 = (2c)^2 - 4a = 4c^2 - 4a
     \]
     Similarly, for \(\Delta_3\) to be a perfect square, \(4c^2 - 4a\) must be a perfect square. Let \(4c^2 - 4a = q^2\) for some integer \(q\). This implies:
     \[
     4(c^2 - a) = q^2 \implies c^2 - a = \left(\frac{q}{2}\right)^2
     \]
     Since \(c, a\) are integers, \(\frac{q}{2}\) must be an integer, so \(q\) must be even. Let \(q = 2r\), then:
     \[
     c^2 - a = r^2 \implies a = c^2 - r^2
     \]

   Now, we have the following system of equations:
   \[
   \begin{cases}
   b = a^2 - m^2 \\
   c = b^2 - p^2 \\
   a = c^2 - r^2
   \end{cases}
   \]

   To find non-negative integer solutions, we can test small values of \(a, b, c\):

   - If \(a = 1\), then:
     \[
     b = 1^2 - 0^2 = 1
     \]
     \[
     c = 1^2 - 0^2 = 1
     \]
     \[
     a = 1^2 - 0^2 = 1
     \]
     Thus, \((a, b, c) = (1, 1, 1)\) is a solution.

2. **Case 2: At least one of \(a, b, c \leq 0\)**

   Since \(a, b, c\) are non-negative integers, one of them must be \(0\). Without loss of generality, let \(a = 0\). This simplifies the first quadratic to:
   \[
   x^2 - 2 \cdot 0 \cdot x + b = 0 \implies x^2 + b = 0
   \]
   For \(x\) to be a non-negative integer, \(b\) must be \(0\). Similarly, the second quadratic simplifies to:
   \[
   x^2 - 2 \cdot 0 \cdot x + c = 0 \implies x^2 + c = 0
   \]
   For \(x\) to be a non-negative integer, \(c\) must be \(0\). Therefore, if one of \(a, b, c \leq 0\), then \(a = b = c = 0\). Thus, \((a, b, c) = (0, 0, 0)\) is another solution.

Conclusion:
The solutions are \((a, b, c) = (1, 1, 1)\) and \((a, b, c) = (0, 0, 0)\).

The final answer is \( \boxed{ (1, 1, 1) } \) and \((0, 0, 0)\).