Lemma 1. Let $A \equiv 2 \pmod{3}$ be a positive integer. Then there exists a prime number $p$ such that $p \equiv 2 \pmod{3}$ and $p^\alpha \mid A$ where $\alpha$ is an odd integer.

Proof of lemma. Assume to the contrary that there is not such a prime number $p$. Therefore, if $A = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ is the prime factorization of $A$, we have two cases for $p_i$'s, $1 \le i \le k$, to consider:

Case 1. $p_i \equiv 1 \pmod{3} \Rightarrow p_i^{\alpha_i} \equiv 1 \pmod{3}$.

Case 2. $p_i \equiv 2 \pmod{3}$ and $2|\alpha_i \Rightarrow p_i^{\alpha_i} \equiv (p_i^2)^{\frac{\alpha_i}{2}} \equiv 1 \pmod{3}$.

As a result, we must have $A \equiv 1 \pmod{3}$ which is a contradiction. $\square$

Lemma 2. Let $p \equiv 2 \pmod{3}$ be a prime number. Show that $\{0^3, 1^3, \dots, (p-1)^3\}$ is a complete residue system modulo $p$.

Proof of lemma. Obviously $i^3 \equiv 0^3 \pmod{p}$ iff $i \equiv 0 \pmod{p}$. Suppose that $p \nmid i, j$. Our goal is to show that $i^3 \equiv j^3 \pmod{p}$ iff $i \equiv j \pmod{p}$. One part of the proof is obvious. To prove the other part, suppose that $p = 3t + 2$. Then by Fermat's Little Theorem, we have $i^{3t+1} \equiv j^{3t+1} \equiv 1 \pmod{p}$. Hence, we have
$$i^{3t} i \equiv i^{3t+1} \equiv j^{3t+1} \equiv (j^3)^t j \equiv i^{3t} j \pmod{p}.$$
Since $(i, p) = 1$, we get $i \equiv j \pmod{p}$. $\square$

Now we are ready to solve the main problem. We claim that there is no such triple. Assume to the contrary that
$$a^2 + b^2 + c^2 = 2013k(ab + bc + ca)$$
for some positive integer $k$.

First, without loss of generality we can suppose that $a$, $b$ and $c$ have no common factor, because if $(a, b, c) = d > 1$, we can divide them by $d$ to get a new triple with no common factor. We have $(a+b+c)^2 = (2013k+2)(ab+bc+ca)$. $2013k+2 \equiv 2 \pmod{3}$, so by lemma 1 there is some prime number $p \equiv 2 \pmod{3}$ such that $p^{2n+1} \nmid 2013k+2$ ($n \ge 0$).
$$\begin{aligned} p^{2n+1} \nmid 2013k+2 & \Rightarrow p^{2n+1}|(a+b+c)^2 \Rightarrow p^{2n+2}|(a+b+c)^2 \\ & \Rightarrow p^{2n+2}|(2013k+2)(ab+bc+ca) \Rightarrow p|ab+bc+ca. \end{aligned}$$
As a result, $p|a+b+c$ and $p|ab+bc+ca$. Hence
$$\begin{aligned} 0 &\equiv ab + bc + ca \equiv ab + c(a+b) \equiv ab + c(-c) \pmod{p} \Rightarrow ab \equiv c^2 \pmod{p} \\ & \Rightarrow c^3 \equiv abc \pmod{p}. \end{aligned}$$
By a similar argument, $a^3 \equiv b^3 \equiv abc \pmod{p}$, so by lemma 2 we deduce $a \equiv b \equiv c \pmod{p}$ and since $p|a+b+c$ and $3 \nmid p$, we find that $p$ divides $a$, $b$ and $c$, which contradicts our assumption that $(a, b, c) = 1$. $\square$