The square of the largest of 3 consecutive integers is 140 less than the sum of the squares of the two smaller integers. find the 3 integers

Let the integers be x-1, x, x+1.

(x+1)² = (x-1)² + x² -140

x² + 2x + 1 = x² - 2x + 1 + x² - 140

2x = -2x + x² - 140

x² -4x -140 = 0

x² -14x +10x -140 = 0

x(x - 14) + 10(x - 14) = 0

(x - 14)(x + 10) = 0

x = 14 or -10

Going back to the criteria mentioned in the question and checking,

--- When x = -10

(-9)² = (-11)² + (-10)² - 140

81 = 121 + 100 - 140

81 = 81

--- When x = 14

(15)² = (13)² + (14)² - 140

225 = 169 + 196 - 140

225 = 225

Since both values satisfy the condition, both values are correct.

Hence, the 3 consecutive integers can be: -11, -10, -9 OR 13, 14, 15.