We claim that the answer is yes. Take $(a,b,c)=(2,7,7)$ Claim 1: $2x^2+7y^2+7z^2$ is never a perfect square. Proof: Work modulo $7$, if $7 \nmid x$ then as $2$ is NQR and $x^2$ is QR so product must be NQR, if $7 \mid x$, then we get that $7(14x^2+y^2+z^2)$ is perfect square, which means $7 \mid y,z$ leading to descent contradiction!. It is well known that $\tau$ is odd only when the input is perfect square, and now Claim 1 proves that we can't construct odd $K$, for even $K$; $K=2$ the construction is $x=2, y = 2, z=13$ which gives $ax+by+cz = 109$; now for $K=2(\ell+1)$ take $A \mapsto A \cdot p^{\ell}$ for $A \in \{2,2,13\}$ giving all even numbers, now as number of even numbers before $2024$ are $1012$, we are done.