How can we solve halting problem efficiently?
The idea is to determine how many of the Turing machines halt. We will show more generally that given $N = 2^n-1$ many Turing machines $T_1,\ldots,T_N$, we can determine which of them halt (on the empty input) using only $n$ calls to an oracle that solves the halting problem.
First, let us see how to determine, using a single oracle call, whether at least $\ell$ of the Turing machines halt. The idea is to construct a Turing machine $S_\ell$ which runs all of $T_1,\ldots,T_N$ in parallel, and halts once $\ell$ of the machines have halted.
Given this, we can determine how many of $T_1,\ldots,T_N$ using binary search. In your case, $N = 3$, this proceeds as follows:

Determine whether at least two of $T_1,T_2,T_3$ halt.
If so, determine whether at least three of $T_1,T_2,T_3$ halt.
Otherwise, determine if at least one of $T_1,T_2,T_3$ halts.

Having found out that exactly $\ell$ of the machines halt, we simply simulate all the machines in parallel until $\ell$ of them halt. The remaining machines will not halt.
It would be interesting to figure out whether this is optimal, that is, whether we can determine the halting of more than $2^n-1$ machines using only $n$ oracle calls.