A three-digit number has digits \(a\), \(b\), and \(c\) such that \(a\) is the hundreds digit, \(b\) is the tens digit, and \(c\) is the units digit. The number can be represented as \(100a + 10b + c\). If the sum of the digits is 12, and the number is three times the sum of the digits when reversed (i.e., when the digits are written in reverse order as \(100c + 10b + a\)), what is the original number? Let's think step by step and output the final answer within \boxed{}.
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