Let n be the smallest of the three consecutive odd integer numbers.
Then the three numbers are n, (n+2) and (n+4).
The equation is
2n + 3*(n+2) + (n+4) = 412
2n + 3n + 6 + n + 4 = 412
6n + 10 = 412
6n = 412 - 10
6n = 402
n = = 67.
ANSWER. The three numbers are 67, 69 and 71.