See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

Find all positive integers $a, b, c$ such that $3abc+11(a+b+c)=6(ab+bc+ac)+18.$
1. **Assume without loss of generality** that \(1 \leq a \leq b \leq c\). This assumption helps to simplify the problem by reducing the number of cases we need to consider.

2. **Rewrite the given equation**:
   \[
   3abc + 11(a + b + c) = 6(ab + bc + ac) + 18
   \]

3. **Isolate \(3abc\)**:
   \[
   3abc = 6(ab + bc + ac) + 18 - 11(a + b + c)
   \]

4. **Analyze the inequality**:
   \[
   3abc < 6(ab + bc + ac) + 18
   \]
   Dividing both sides by 3:
   \[
   abc < 2(ab + bc + ac) + 6
   \]

5. **Use the assumption \(1 \leq a \leq b \leq c\)** to further simplify**:
   \[
   a < 2\left(\frac{a}{c} + 1 + \frac{a}{b}\right) + \frac{6}{bc}
   \]
   Since \(1 \leq a \leq b \leq c\), we have:
   \[
   \frac{a}{c} \leq 1 \quad \text{and} \quad \frac{a}{b} \leq 1
   \]
   Therefore:
   \[
   a < 2(1 + 1 + 1) + 6 = 12
   \]
   This implies:
   \[
   a \leq 12
   \]

6. **Consider the divisibility condition**:
   Since \(a + b + c\) should be divisible by 3, we can set \(a + b + c = 3k\) for some positive integer \(k\).

7. **Rewrite the equation using \(k\)**:
   \[
   abc = 2(ab + bc + ac) + 6 - 11k
   \]

8. **Check for small values of \(a\)**:
   - For \(a = 1\):
     \[
     3bc + 11(1 + b + c) = 6(b + c + bc) + 18
     \]
     Simplifying:
     \[
     3bc + 11 + 11b + 11c = 6bc + 6b + 6c + 18
     \]
     \[
     3bc + 11b + 11c + 11 = 6bc + 6b + 6c + 18
     \]
     \[
     3bc + 5b + 5c + 11 = 6bc + 18
     \]
     \[
     3bc - 6bc + 5b + 5c + 11 = 18
     \]
     \[
     -3bc + 5b + 5c + 11 = 18
     \]
     \[
     -3bc + 5b + 5c = 7
     \]
     \[
     3bc = 5b + 5c - 7
     \]
     Solving for \(b\) and \(c\) under the condition \(1 \leq b \leq c\), we find that \(b = 1\) and \(c = 2\) works:
     \[
     3(1)(2) = 5(1) + 5(2) - 7
     \]
     \[
     6 = 5 + 10 - 7
     \]
     \[
     6 = 8
     \]
     This is not true, so we need to check other values.

9. **Continue checking for other small values of \(a\)**:
   - For \(a = 2\), \(a = 3\), etc., until we find a valid solution.

10. **Verify the solution**:
    After checking, we find that \(a = 1\), \(b = 2\), and \(c = 3\) satisfy the equation:
    \[
    3(1)(2)(3) + 11(1 + 2 + 3) = 6(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 1) + 18
    \]
    \[
    18 + 66 = 6(2 + 6 + 3) + 18
    \]
    \[
    84 = 6 \cdot 11 + 18
    \]
    \[
    84 = 66 + 18
    \]
    \[
    84 = 84
    \]

The final answer is \( \boxed{ (a, b, c) = (1, 2, 3) } \)