With three different digits, all greater than $0$, six different three-digit numbers are formed. If we add these six numbers together the result is $4.218$. The sum of the three largest numbers minus the sum of the three smallest numbers equals $792$. Find the three digits.
1. Let the three different digits be \(A\), \(B\), and \(C\). The six different three-digit numbers that can be formed are:
   \[
   100A + 10B + C, \quad 100A + 10C + B, \quad 100B + 10A + C, \quad 100B + 10C + A, \quad 100C + 10A + B, \quad 100C + 10B + A
   \]

2. Adding these six numbers together, we get:
   \[
   (100A + 10B + C) + (100A + 10C + B) + (100B + 10A + C) + (100B + 10C + A) + (100C + 10A + B) + (100C + 10B + A)
   \]
   Simplifying, we have:
   \[
   2(100A + 100B + 100C) + 2(10A + 10B + 10C) + 2(A + B + C) = 222(A + B + C)
   \]
   Given that the sum of these six numbers is 4218, we can write:
   \[
   222(A + B + C) = 4218
   \]
   Solving for \(A + B + C\):
   \[
   A + B + C = \frac{4218}{222} = 19
   \]

3. Without loss of generality, assume \(A > B > C\). We are also given that the sum of the three largest numbers minus the sum of the three smallest numbers equals 792. The three largest numbers are:
   \[
   100A + 10B + C, \quad 100A + 10C + B, \quad 100B + 10A + C
   \]
   The three smallest numbers are:
   \[
   100B + 10C + A, \quad 100C + 10A + B, \quad 100C + 10B + A
   \]
   The difference between the sum of the three largest and the sum of the three smallest is:
   \[
   (100A + 10B + C) + (100A + 10C + B) + (100B + 10A + C) - (100B + 10C + A) - (100C + 10A + B) - (100C + 10B + A)
   \]
   Simplifying, we get:
   \[
   198A - 198C = 792
   \]
   Solving for \(A - C\):
   \[
   198(A - C) = 792 \implies A - C = \frac{792}{198} = 4
   \]

4. We now have two equations:
   \[
   A + B + C = 19
   \]
   \[
   A - C = 4
   \]
   Solving these equations, we substitute \(A = C + 4\) into the first equation:
   \[
   (C + 4) + B + C = 19 \implies 2C + B + 4 = 19 \implies 2C + B = 15 \implies B = 15 - 2C
   \]

5. Since \(A\), \(B\), and \(C\) are digits greater than 0, we test possible values for \(C\):
   - If \(C = 3\):
     \[
     B = 15 - 2(3) = 9 \quad \text{and} \quad A = 3 + 4 = 7
     \]
     This gives \(A = 7\), \(B = 9\), \(C = 3\). However, \(A > B > C\) is not satisfied.
   - If \(C = 4\):
     \[
     B = 15 - 2(4) = 7 \quad \text{and} \quad A = 4 + 4 = 8
     \]
     This gives \(A = 8\), \(B = 7\), \(C = 4\). Here, \(A > B > C\) is satisfied.
   - If \(C = 5\):
     \[
     B = 15 - 2(5) = 5 \quad \text{and} \quad A = 5 + 4 = 9
     \]
     This gives \(A = 9\), \(B = 5\), \(C = 5\). However, \(A > B > C\) is not satisfied.

Thus, the only valid solution is \(A = 8\), \(B = 7\), \(C = 4\).

The final answer is \(\boxed{8, 7, 4}\).