Find all the triples of integers $ (a, b,c)$ such that:
\[ \begin{array}{ccc}a\plus{}b\plus{}c &\equal{}& 24\\ a^{2}\plus{}b^{2}\plus{}c^{2}&\equal{}& 210\\ abc &\equal{}& 440\end{array}\]
1. We start with the given equations:
   \[
   a + b + c = 24
   \]
   \[
   a^2 + b^2 + c^2 = 210
   \]
   \[
   abc = 440
   \]

2. First, we square the sum of \(a\), \(b\), and \(c\):
   \[
   (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
   \]
   Substituting the known values:
   \[
   24^2 = 210 + 2(ab + bc + ca)
   \]
   \[
   576 = 210 + 2(ab + bc + ca)
   \]
   Solving for \(ab + bc + ca\):
   \[
   576 - 210 = 2(ab + bc + ca)
   \]
   \[
   366 = 2(ab + bc + ca)
   \]
   \[
   ab + bc + ca = 183
   \]

3. We now use the relationships \(ab = \frac{440}{c}\) and \(a + b = 24 - c\). Substituting these into the equation for \(ab + bc + ca\):
   \[
   \frac{440}{c} + c(24 - c) = 183
   \]
   Simplifying:
   \[
   440 + c^2(24 - c) = 183c
   \]
   \[
   440 + 24c^2 - c^3 = 183c
   \]
   Rearranging into a standard polynomial form:
   \[
   c^3 - 24c^2 + 183c - 440 = 0
   \]

4. We solve the cubic equation \(c^3 - 24c^2 + 183c - 440 = 0\) by factoring:
   \[
   (c - 5)(c - 8)(c - 11) = 0
   \]
   Thus, the possible values for \(c\) are \(5\), \(8\), and \(11\).

5. For each value of \(c\), we solve for \(a\) and \(b\):

   - **Case 1: \(c = 5\)**
     \[
     a + b = 24 - 5 = 19
     \]
     \[
     ab = \frac{440}{5} = 88
     \]
     Solving the quadratic equation \(t^2 - 19t + 88 = 0\):
     \[
     t = \frac{19 \pm \sqrt{19^2 - 4 \cdot 88}}{2} = \frac{19 \pm \sqrt{361 - 352}}{2} = \frac{19 \pm 3}{2}
     \]
     \[
     t = 11 \quad \text{or} \quad t = 8
     \]
     Thus, the pairs \((a, b)\) are \((11, 8)\) and \((8, 11)\).

   - **Case 2: \(c = 8\)**
     \[
     a + b = 24 - 8 = 16
     \]
     \[
     ab = \frac{440}{8} = 55
     \]
     Solving the quadratic equation \(t^2 - 16t + 55 = 0\):
     \[
     t = \frac{16 \pm \sqrt{16^2 - 4 \cdot 55}}{2} = \frac{16 \pm \sqrt{256 - 220}}{2} = \frac{16 \pm 6}{2}
     \]
     \[
     t = 11 \quad \text{or} \quad t = 5
     \]
     Thus, the pairs \((a, b)\) are \((11, 5)\) and \((5, 11)\).

   - **Case 3: \(c = 11\)**
     \[
     a + b = 24 - 11 = 13
     \]
     \[
     ab = \frac{440}{11} = 40
     \]
     Solving the quadratic equation \(t^2 - 13t + 40 = 0\):
     \[
     t = \frac{13 \pm \sqrt{13^2 - 4 \cdot 40}}{2} = \frac{13 \pm \sqrt{169 - 160}}{2} = \frac{13 \pm 3}{2}
     \]
     \[
     t = 8 \quad \text{or} \quad t = 5
     \]
     Thus, the pairs \((a, b)\) are \((8, 5)\) and \((5, 8)\).

6. Combining all the solutions, we get the following triples \((a, b, c)\):
   \[
   \{(11, 8, 5), (8, 11, 5), (8, 5, 11), (5, 8, 11), (11, 5, 8), (5, 11, 8)\}
   \]

The final answer is \( \boxed{ \{(11, 8, 5), (8, 11, 5), (8, 5, 11), (5, 8, 11), (11, 5, 8), (5, 11, 8)\} } \)