Solve the equation $$x^2+y^2+z^2=686$$ where $x$, $y$ and $z$ are positive integers
To solve the equation \( x^2 + y^2 + z^2 = 686 \) where \( x \), \( y \), and \( z \) are positive integers, we can proceed as follows:

1. **Assume \( x \leq y \leq z \)**:
   This assumption helps reduce the number of cases we need to check by considering only the ordered triplets.

2. **Determine the upper bound for \( x \)**:
   Since \( x \leq y \leq z \), we can find an upper bound for \( x \) by considering the maximum possible value for \( x \). If \( x \) were too large, \( y \) and \( z \) would not be able to satisfy the equation. We can start by noting that:
   \[
   x^2 \leq \frac{686}{3} \approx 228.67 \implies x \leq \sqrt{228.67} \approx 15.12
   \]
   Therefore, \( x \leq 15 \).

3. **Check all possible values of \( x \) from 1 to 15**:
   For each value of \( x \), we need to find pairs \((y, z)\) such that \( y \leq z \) and \( x^2 + y^2 + z^2 = 686 \).

4. **Iterate through possible values of \( y \) and \( z \)**:
   For each \( x \), iterate through possible values of \( y \) and \( z \) such that \( y \leq z \) and check if they satisfy the equation.

Let's go through the steps for each \( x \):

- For \( x = 1 \):
  \[
  1^2 + y^2 + z^2 = 686 \implies y^2 + z^2 = 685
  \]
  Check pairs \((y, z)\):
  - \( y = 3 \), \( z = 26 \) since \( 3^2 + 26^2 = 9 + 676 = 685 \)
  - \( y = 18 \), \( z = 19 \) since \( 18^2 + 19^2 = 324 + 361 = 685 \)

- For \( x = 5 \):
  \[
  5^2 + y^2 + z^2 = 686 \implies y^2 + z^2 = 661
  \]
  Check pairs \((y, z)\):
  - \( y = 6 \), \( z = 25 \) since \( 6^2 + 25^2 = 36 + 625 = 661 \)

- For \( x = 6 \):
  \[
  6^2 + y^2 + z^2 = 686 \implies y^2 + z^2 = 650
  \]
  Check pairs \((y, z)\):
  - \( y = 11 \), \( z = 23 \) since \( 11^2 + 23^2 = 121 + 529 = 650 \)
  - \( y = 17 \), \( z = 19 \) since \( 17^2 + 19^2 = 289 + 361 = 650 \)

- For \( x = 7 \):
  \[
  7^2 + y^2 + z^2 = 686 \implies y^2 + z^2 = 637
  \]
  Check pairs \((y, z)\):
  - \( y = 14 \), \( z = 21 \) since \( 14^2 + 21^2 = 196 + 441 = 637 \)

- For \( x = 9 \):
  \[
  9^2 + y^2 + z^2 = 686 \implies y^2 + z^2 = 605
  \]
  Check pairs \((y, z)\):
  - \( y = 11 \), \( z = 22 \) since \( 11^2 + 22^2 = 121 + 484 = 605 \)

- For \( x = 10 \):
  \[
  10^2 + y^2 + z^2 = 686 \implies y^2 + z^2 = 586
  \]
  Check pairs \((y, z)\):
  - \( y = 15 \), \( z = 19 \) since \( 15^2 + 19^2 = 225 + 361 = 586 \)

5. **List all solutions**:
   The solutions are:
   \[
   (x, y, z) = (1, 3, 26), (1, 18, 19), (5, 6, 25), (6, 11, 23), (6, 17, 19), (7, 14, 21), (9, 11, 22), (10, 15, 19)
   \]
   and their permutations.

The final answer is \( \boxed{ (1, 3, 26), (1, 18, 19), (5, 6, 25), (6, 11, 23), (6, 17, 19), (7, 14, 21), (9, 11, 22), (10, 15, 19) } \) and their permutations.