It given three cases
Case I : When f(x) = 1 is true.
In this case remaining two are false
∴f(y) = 1 and f(z) = 2
This means x and y have the same image, so f(x) is not an injective, which is a contradiction.
Case II : when f(y) ≠ 1 is true.
If f(y) ≠ 1 is true then the remaining statement are false;
∴ f(x) ≠ 1 and f(z) = 2
ie, both x and y are not mapped to 1. so, either both associate to 2 or 3, Thus, it is not injective.
Case III : When f(z) ≠ 2 is true
If f(z) ≠ 2 is true then remaining statement are false
∴ If f(x) ≠ 1 and f(y) = 1
But f is injective
Thus, we have f(x) = 2, f(y) = 1 and f(z) = 3
Hence, f-1(1) = y