Determine all triples $(a, b, c)$ of nonnegative integers that satisfy: $$(c-1) (ab- b -a) = a + b-2$$
We start with the given equation:
\[
(c-1)(ab - b - a) = a + b - 2
\]

First, we rearrange the equation to isolate terms involving \(a\) and \(b\):
\[
(c-1)ab - (c-1)b - (c-1)a = a + b - 2
\]

Next, we distribute and rearrange:
\[
(c-1)ab - (c-1)b - (c-1)a - a - b = -2
\]

Combine like terms:
\[
(c-1)ab - (c-1 + 1)a - (c-1 + 1)b = -2
\]

Simplify:
\[
(c-1)ab - ca - cb + a + b = -2
\]

Rearrange to group terms involving \(a\) and \(b\):
\[
(c-1)ab - ca - cb + a + b = -2
\]

We can rewrite this as:
\[
(c-1)ab - ca - cb + a + b + 2 = 0
\]

Now, we need to find non-negative integer solutions to this equation. We will consider different cases for \(c\).

**Case 1: \(c = 0\)**

Substitute \(c = 0\) into the equation:
\[
(-1)ab - 0a - 0b + a + b + 2 = 0
\]
\[
-ab + a + b + 2 = 0
\]
\[
a + b - ab + 2 = 0
\]
\[
ab - a - b = 2
\]

We need to find non-negative integer solutions to this equation. Testing small values, we find:
\[
(a, b) = (2, 1), (1, 2)
\]

Thus, the solutions for \(c = 0\) are:
\[
(2, 1, 0), (1, 2, 0)
\]

**Case 2: \(c = 1\)**

Substitute \(c = 1\) into the equation:
\[
(1-1)ab - 1a - 1b + a + b + 2 = 0
\]
\[
0 - a - b + a + b + 2 = 0
\]
\[
2 = 0
\]

This is a contradiction, so there are no solutions for \(c = 1\).

**Case 3: \(c = 2\)**

Substitute \(c = 2\) into the equation:
\[
(2-1)ab - 2a - 2b + a + b + 2 = 0
\]
\[
ab - 2a - 2b + a + b + 2 = 0
\]
\[
ab - a - b = 2
\]

We need to find non-negative integer solutions to this equation. Testing small values, we find:
\[
(a, b) = (3, 4), (4, 3), (1, 0), (0, 1)
\]

Thus, the solutions for \(c = 2\) are:
\[
(3, 4, 2), (4, 3, 2), (1, 0, 2), (0, 1, 2)
\]

**Case 4: \(c = 3\)**

Substitute \(c = 3\) into the equation:
\[
(3-1)ab - 3a - 3b + a + b + 2 = 0
\]
\[
2ab - 3a - 3b + a + b + 2 = 0
\]
\[
2ab - 2a - 2b = 2
\]
\[
ab - a - b = 1
\]

We need to find non-negative integer solutions to this equation. Testing small values, we find:
\[
(a, b) = (2, 4), (4, 2)
\]

Thus, the solutions for \(c = 3\) are:
\[
(2, 4, 3), (4, 2, 3)
\]

**Case 5: \(c = 4\)**

Substitute \(c = 4\) into the equation:
\[
(4-1)ab - 4a - 4b + a + b + 2 = 0
\]
\[
3ab - 4a - 4b + a + b + 2 = 0
\]
\[
3ab - 3a - 3b = 2
\]
\[
ab - a - b = \frac{2}{3}
\]

Since \(\frac{2}{3}\) is not an integer, there are no solutions for \(c = 4\).

Combining all the solutions, we have:
\[
(2, 1, 0), (1, 2, 0), (3, 4, 2), (4, 3, 2), (1, 0, 2), (0, 1, 2), (2, 4, 3), (4, 2, 3)
\]

The final answer is \( \boxed{ (2, 1, 0), (1, 2, 0), (3, 4, 2), (4, 3, 2), (1, 0, 2), (0, 1, 2), (2, 4, 3), (4, 2, 3) } \).