The correct option is B 10, 11, 12
Let the three consecutive positive integers be x, x+1 and x+2.
Given that, sum of square of the first number and the product of other two numbers is 232
⇒x2+(x+1)(x+2)=232
⇒x2+(x2+3x+2)=232
⇒2x2+3x−230=0
Using quadratic formula, x=−b±√b2−4ac2a
Here a=2,b=3 and c=−230
⇒x=−3±√32−4×2×−2304
=−3±√18494
=−3±434
⇒x=10 or −11.5
Ignoring negative value, since x is a positive integer.
∴ x = 10.
Hence, the numbers are 10, 11 and 12.