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.