Let f(x) = x^{3} - 6x^{2} + 11x - 6

Given: (x-1)(x-2) are the zeros of f(x).

product of zeros = (x-1)(x-2) = x^{2} - 2x - x + 2 = x^{2} -3x + 2

therefore, on dividing the f(x) with x^{2} - 3x + 2 , we get, x-3

Now, x - 3 = 0

x = 3

Hence, the other zero is 3.