Find three consecutive odd integers such that the sum of one times the first, three times the second, and two times the third is -232. List the numbers in order from smallest to largest.
Let's represent the three consecutive odd integers as $n$, $n+2$, and $n+4$.
According to the given information, we can set up the equation $1n + 3(n+2) + 2(n+4) = -232$, and simplify it as follows:
$1n + 3(n+2) + 2(n+4) = -232$
$1n + 3n + 6 + 2n + 8 = -232$
$6n + 14 = -232$
$6n = -246$
$n = -41$
Therefore, the three consecutive odd integers are $-41$, $-39$, and $-37$.