Three, (Full marks 10 points) The numbers A, B, and C are 312, $270$, and 211, respectively. When these three numbers are divided by a natural number $A$, the remainder of dividing A is twice the remainder of dividing B, and the remainder of dividing B is twice the remainder of dividing C. Find this natural number $A$.

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Note: In the problem, the numbers are referred to as 甲 (A), 乙 (B), and 丙 (C), but for clarity in English, I have used A, B, and C to represent the numbers 312, 270, and 211, respectively.
Three, from the problem, we have
$$
270 \times 2-312=228, \quad 211 \times 2-270=152 \text {. }
$$

Then we have
\begin{tabular}{|c|c|c|}
\hline \multicolumn{2}{|c|}{$2 \mid \quad 228$} & 152 \\
\hline $2 T$ & 114 & 76 \\
\hline 19 & 57 & 38 \\
\hline & 3 & 2 \\
\hline
\end{tabular}

Therefore, 4 could be $2, 4, 19, 38, 19 \times 2^{2}$.
Upon verification, only 19 meets the requirement.