Let three consecutive integers be x, x + 1, x + 2. According to the question,
2x + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
On transposing 11 to R.H.S, we obtain
9x = 74 − 11
9x = 63
On dividing both sides by 9, we obtain
9x/9 = 63/9
x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
Hence, the numbers are 7, 8, and 9.