That numerator doesn't impose any restrictions, but there are two constraints from the denominator: the whole denominator cannot be 0 and the number inside the radical cannot be negative. Let's look at each in turn.

The denominator cannot be 0 So then we cannot have \(\displaystyle{3}−\sqrt{x}−{2}={0}\). So let's find out which x's make this true so that we can exclude them from the domain.

\(\displaystyle{3}=\sqrt{{{x}-{2}}}\)

\(9=x-2\)

\(11=x\)

So 11 cannot be in the domain.

The number under the radical cannot be negative. That is, we must have \(\displaystyle{x}−{2}≥{0}\). Hence \(\displaystyle{x}≥{2}\).

Conclusion

Putting these two conditions together we get that the domain is \(\displaystyle{\left[{2},∞\right)}\) without the number 11. Hence

Domain \(\displaystyle{\left({f}\right)}={\left[{2},{11}\right)}∪{\left({11},∞\right)}\)