The answer is yes. Take any triple $(a,b,c)$ with ${\rm gcd}(a,b,c)=1$. I prove that the set \[ \left\{v_2\left(a^k+b^k+c^k\right): k\in\mathbb{N}\right\} \]is finite. This will yield the claim. We argue by parity. Note that not all $a,b,c$ are even. If exactly two of them is even or none is even then $a^k+b^k+c^k$ is odd for any $k$, so it suffices to let $n=1$. Lastly, assume exactly one of them even: let $a,b$ be odd and $2\mid c$. Note that if $k$ is even then $a^k+b^k\equiv 2\pmod{4}$, so $v_2(a^k+b^k)=1$. If $k$ is odd, then $v_2(a^k+b^k)=v_2(a+b)$. So, for any $k>K_1 = v_2(a+b)/v_2(c)$, we get $v_2(a^k+b^k+c^k) = v_2(a+b)$, yielding the claim.