Find three consecutive odd integers such that three times the square of the first integer is twelve more than the product of the second and the third integers

Consecutive odd integers are 2 apart

they are

n, n+2, n+4

3 (n²) = 12 + (n+2) (n+4)

expand

3n²=12+n²+6n+8

3n²=n²+6n+20

minus n² both sides

2n²=6n+20

divide bot sides by 2

n²=3n+10

minus 3n+10 both sides

n²-3n-10=0

factor

(n-5) (n+2) = 0

n-5=0

n=5

n+2=0

n=-2

has to be odd, so no

n=5

n+2=7

n+4=9

the integers are 5,7,9