Solution:

The answer is no: such integers do not exist. In what follows, $\nu_{p}(n)$ will denote the exponent of $p$ in the prime factorization of $n$.

Assume without loss of generality that $a, b, c$ do not have some common divisor, and $a^{2} + b^{2} + c^{2} = 3k(a b + b c + c a)$. Write
$$(3k + 2)\left(a^{2} + b^{2} + c^{2}\right) = 3k \cdot (a + b + c)^{2}.$$

Since $3k + 2 \equiv 2 \pmod{3}$, there is a prime $p \equiv 2 \pmod{3}$ with $\nu_{p}(3k + 2)$ odd (in particular, $p \mid 3k + 2$).
We first show that $p \neq 2$. Let us assume on the contrary that $\nu_{2}(3k + 2)$ is odd (in particular, $k$ is even). Remark that since $a, b, c$ are not all even, $\nu_{2}\left(a^{2} + b^{2} + c^{2}\right) \leq 1$. Furthermore,
$$\nu_{2}(a + b + c) = 0 \Longleftrightarrow \nu_{2}\left(a^{2} + b^{2} + c^{2}\right) = 0.$$

Now we consider two cases.
- Assume $\nu_{2}(k) \geq 2$. Then $\nu_{2}(3k + 2) = 1$, $\nu_{2}\left(a^{2} + b^{2} + c^{2}\right) \leq 1$. Therefore
$$\nu_{2}(k) + 2\nu_{2}(a + b + c) \geq 2 \geq \nu_{2}(3k + 2) + \nu_{2}\left(a^{2} + b^{2} + c^{2}\right)$$
but equality cannot occur since the relations
$$\nu_{2}(a + b + c) = 0 \quad \text{and} \quad \nu_{2}\left(a^{2} + b^{2} + c^{2}\right) = 1$$
cannot hold simultaneously.

- Assume $\nu_{2}(k) = 1$. Then $\nu_{2}(3k + 2) > 1$ and is odd. Now
$$\nu_{2}(3k + 2) + \nu_{2}\left(a^{2} + b^{2} + c^{2}\right) = 1 + 2\nu_{2}(a + b + c).$$
Thus $\nu_{2}\left(a^{2} + b^{2} + c^{2}\right)$ must be even, so it is zero; consequently $\nu_{2}(a + b + c) = 0$ as well and we obtain $1 < v_{2}(3k + 2) = 1$.

Now for the interesting part. Remark $p \mid a + b + c$ and $p \mid a^{2} + b^{2} + c^{2}$. Without loss of generality $b \not\equiv 0 \pmod{p}$, so that
$$a^{2} + b^{2} + (a + b)^{2} \equiv 0 \pmod{p} \Longrightarrow a^{2} + a b + b^{2} \equiv 0 \pmod{p}$$
Then if $x = a b^{-1}$, we get that $x^{2} + x + 1 \equiv 0 \pmod{p}$. But the left-hand side is the third cyclotomic polynomial, so either $p = 3$ or $3 \mid p - 1$, but neither is the case.

Therefore, such a triple $(a, b, c)$ does not exist.