Determine all integers $a, b, c$ satisfying identities:
$a + b + c = 15$
$(a - 3)^3 + (b - 5)^3 + (c -7)^3 = 540$
To determine all integers \(a, b, c\) satisfying the identities \(a + b + c = 15\) and \((a - 3)^3 + (b - 5)^3 + (c - 7)^3 = 540\), we will proceed step-by-step.

1. **Substitute new variables**:
   Let \(x = a - 3\), \(y = b - 5\), and \(z = c - 7\). Then the given equations transform as follows:
   \[
   (x + 3) + (y + 5) + (z + 7) = 15
   \]
   Simplifying, we get:
   \[
   x + y + z + 15 = 15 \implies x + y + z = 0
   \]
   The second equation becomes:
   \[
   x^3 + y^3 + z^3 = 540
   \]

2. **Use the identity for the sum of cubes**:
   Since \(x + y + z = 0\), we can use the identity:
   \[
   x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
   \]
   Given \(x + y + z = 0\), the identity simplifies to:
   \[
   x^3 + y^3 + z^3 = 3xyz
   \]
   Therefore, we have:
   \[
   3xyz = 540 \implies xyz = 180
   \]

3. **Solve for \(x, y, z\)**:
   We need to find integer solutions for \(x, y, z\) such that \(x + y + z = 0\) and \(xyz = 180\).

4. **Factor pairs of 180**:
   Consider the factor pairs of 180. We need to find combinations that sum to zero. The factor pairs of 180 are:
   \[
   (1, 180), (2, 90), (3, 60), (4, 45), (5, 36), (6, 30), (9, 20), (10, 18), (12, 15)
   \]
   We need to find combinations of these factors that sum to zero when considering their signs.

5. **Check possible combinations**:
   - For \(x + y + z = 0\) and \(xyz = 180\):
     - \(x = 9, y = -5, z = -4\)
     - \(x = -5, y = 9, z = -4\)
     - \(x = -4, y = -5, z = 9\)
     - \(x = -5, y = -4, z = 9\)

6. **Convert back to \(a, b, c\)**:
   Using \(x = a - 3\), \(y = b - 5\), and \(z = c - 7\):
   - For \(x = 9, y = -5, z = -4\):
     \[
     a = x + 3 = 12, \quad b = y + 5 = 0, \quad c = z + 7 = 3
     \]
   - For \(x = -5, y = 9, z = -4\):
     \[
     a = x + 3 = -2, \quad b = y + 5 = 14, \quad c = z + 7 = 3
     \]
   - For \(x = -4, y = -5, z = 9\):
     \[
     a = x + 3 = -1, \quad b = y + 5 = 0, \quad c = z + 7 = 16
     \]
   - For \(x = -5, y = -4, z = 9\):
     \[
     a = x + 3 = -2, \quad b = y + 5 = 1, \quad c = z + 7 = 16
     \]

The final answer is \(\boxed{(a, b, c) = (12, 0, 3), (-2, 14, 3), (-1, 0, 16), (-2, 1, 16)}\).