Pick three $a,b,c$ vectors in $\mathbb Z^n$ uniformly with $\max(\|a\|_\infty,\|b\|_\infty)<T$ and $\|c\|_\infty<T^2$ and an $\epsilon>0$.

Assume $a$ and $b$ are coordinatewise coprime (that is every $a_i$ and $b_i$ are coprime at every $i\in\{1,\dots,n\}$). Then do we always have such $A$ and $B$ of absolute value $O(T)$ at general $n$ such that $\|Aa+Bb-c\|_\infty<T^{2(n-1)/n+\epsilon}$ at large enough $n$?

This is the intuition.

Essentially there are $T^2$ choices for $A,B$ and there are $T^{\frac{(2n−2)}n+\epsilon}$ choices for every coordinate of $Aa+Bb-c$ and since there are $n$ coordinates we have $T^2T^{{2n−2}+n\epsilon}=T^{2n+n\epsilon}$ choices. However typically $Aa+Bb-c$ is of size $T^2$ and so typically there are $T^{2n}$ choices. If $\epsilon>0$ then the heuristic that $$\frac{\{\mbox{number of choices for A,B}\}\times\{\mbox{number of vectors with }\infty\mbox{ norm GAP < }T^{\frac{2(n-1)}n+\epsilon}\}}{\mbox{number of length n vectors with coordinates of size }T^2}$$ $$\asymp\frac{T^2T^{(\frac{2(n-1)}n+\epsilon)n}}{T^{2n}}=T^{n\epsilon}$$ holds which is at least $1$ if $\epsilon>0$ implies the $\infty$ norm bound looks plausible.

Also the original problem was what is the probability that the bound holds for uniformly random vectors $a,b,c$ in $\mathbb Z^n$ with $\max(\|a\|_\infty,\|b\|_\infty)<T$ and $\|c\|_\infty<T^2$ and an $\epsilon>0$?

At $n=1$ we get $\frac6{\pi^2}$.

**Update** The original writeup had a miscalculation. I missed a $2$ in exponent (that made the problem silly which some poster answered) and a related $T^2$ part (it was very clear from the denominator that the heuristic assumed a $T^2$ part on $c$ and I had not written it down in post) and it became WWIII and the accepted answer was for that original problem.