To find the domain of the function $f\left(x\right)=\sqrt{x-2}+\sqrt{3-x}$, we need to ensure that the expressions inside the square roots are non-negative. Therefore, we set up the following system of inequalities:

1. For $\sqrt{x-2}$, we have:
$$x-2 \geqslant 0$$
Solving for $x$, we find:
$$x \geqslant 2$$

2. For $\sqrt{3-x}$, we have:
$$3-x \geqslant 0$$
Solving for $x$, we find:
$$x \leqslant 3$$

Combining these two inequalities, we get the intersection of the two solution sets:
$$2 \leqslant x \leqslant 3$$

Therefore, the domain of the function is the interval where both conditions are satisfied, which is:
$$\boxed{\left[2, 3\right]}$$