If $x$, $y$, and $z$ are integers satisfying $xyz+4(x+y+z)=2(xy+xz+yz)+7$, list all possibilities for the ordered triple $(x, y, z)$.
1. Start with the given equation:
   \[
   xyz + 4(x + y + z) = 2(xy + xz + yz) + 7
   \]

2. Rearrange the equation to group similar terms:
   \[
   xyz + 4x + 4y + 4z = 2xy + 2xz + 2yz + 7
   \]

3. Subtract \(2xy + 2xz + 2yz\) from both sides:
   \[
   xyz + 4x + 4y + 4z - 2xy - 2xz - 2yz = 7
   \]

4. Rewrite the left-hand side to factor it:
   \[
   xyz - 2xy - 2xz - 2yz + 4x + 4y + 4z = 7
   \]

5. Notice that the left-hand side can be factored as:
   \[
   (x-2)(y-2)(z-2) = -1
   \]

6. Solve for the integer solutions of \((x-2)(y-2)(z-2) = -1\). The possible factorizations of \(-1\) into three integers are:
   \[
   (1, 1, -1), (1, -1, 1), (-1, 1, 1), (-1, -1, -1)
   \]

7. Convert these factorizations back to the original variables \(x, y, z\):
   - For \((1, 1, -1)\):
     \[
     (x-2, y-2, z-2) = (1, 1, -1) \implies (x, y, z) = (3, 3, 1)
     \]
   - For \((1, -1, 1)\):
     \[
     (x-2, y-2, z-2) = (1, -1, 1) \implies (x, y, z) = (3, 1, 3)
     \]
   - For \((-1, 1, 1)\):
     \[
     (x-2, y-2, z-2) = (-1, 1, 1) \implies (x, y, z) = (1, 3, 3)
     \]
   - For \((-1, -1, -1)\):
     \[
     (x-2, y-2, z-2) = (-1, -1, -1) \implies (x, y, z) = (1, 1, 1)
     \]

8. Therefore, the possible ordered triples \((x, y, z)\) are:
   \[
   (1, 1, 1), (3, 3, 1), (3, 1, 3), (1, 3, 3)
   \]

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