To solve this problem, we need to find all triples of positive integers $(a, b, c)$ such that:

1. $a + b + c = 3k + 1$,
2. $ab + bc + ca = 3k^2 + 2k$.

First, we observe that the problem is set with symmetric conditions which often suggest that $(a, b, c)$ could take a symmetric form. Therefore, let's assume $a = k+1$, $b = k$, and $c = k$. We need to verify that this satisfies both equations.

### Step 1: Verify the First Equation

For the first equation:
$$a + b + c = (k+1) + k + k = 3k + 1.$$
This matches the given condition $a + b + c = 3k + 1$.

### Step 2: Verify the Second Equation

Now, for the second equation:
$$ab + bc + ca = (k+1)k + k(k) + k(k+1).$$

Calculate the individual terms:
- $(k+1)k = k^2 + k$,
- $k^2$,
- $k(k+1) = k^2 + k$.

Add them up:
$$ab + bc + ca = (k^2 + k) + k^2 + (k^2 + k) = 3k^2 + 2k.$$
This also matches the given condition $ab + bc + ca = 3k^2 + 2k$.

### Conclusion:

The symmetric form $(a, b, c) = (k+1, k, k)$ satisfies both conditions of the problem. Therefore, the only solution for the triples $(a, b, c)$ is:
$$\boxed{(k+1, k, k)}.$$

We conclude that the solution to the problem is $(k+1, k, k)$ based on the given constraints and conditions.