The answer is all even positive integers. To achieve $k=2l$ take $(a,b)=(2\cdot 5^l,3\cdot 5^l)$ so that $d(2\cdot 5^l)=d(3\cdot 5^l)=d(13\cdot 5^l)=2(l+1)$ with $l\geq 0$. If $k$ was odd, all three numbers would be squares, and so $a=m^2$, $b=n^2$ and $c=2m^2+3n^2=p^2$. Taking mod 3 we have $2m^2\equiv p^2$ so $m\equiv p\equiv 0\pmod{3}$. Therefore $9|p^2-2m^2=3n^2$, so $3|n$. Thus, by infinite descent, $m=n=p=0$ which of course isn't allowed. Therefore the only $k$s achievable are the even positive integers