The original equation states
$\frac{(ab)^2 + 3ab - 18 }{ (a-1)(a+2)}$

The first thing to note is the denominator. For the given expression to be true

$\implies a \neq 1$ and $a \neq -2$
as these a =1 or a = -2 will result in an undefined expression irrespective of the value of b.

Since the numerator is a quadratic expression in the variables ab, lets replace ab with x.


Factorizing




Since the possible values of b include only 1, 2 and 3, therefore we will focus only on these factors of b.

For $ab =3 \implies a =1 \ \ b = 3$ or $a = 3, b = 1$
However since $a \neq 1$, only the second option i.e. b = 1 offers a valid choice. Thus I. is valid.

For
$ab =-6 \implies a = -6, b = 1$ or $a = -3, b = 2$ or $a = -2,\ b = 3$

The first choice for b is already valid i.e. b = 1 and does not invalidate the conditions for a. The second choice i.e b = 2 is also valid, thus II. is true. However, the third option is invalid since it requires a = -2 which we understand is an invalid choice for a

Option C states both I and II and thus Option C is the right answer.