The domain of $$f(x) = \frac{\cos^{-1}\left(\frac{x^2 - 5x + 6}{x^2 - 9}\right)}{\log(x^2 - 3x + 2)}$$ is

Given,

$$f(x)=\frac{\cos^{-1}\left(\dfrac{x^2-5x+6}{x^2-9}\right)}{\log(x^2-3x+2)}$$

For the function to be defined, we need:

1. $$-1\le \dfrac{x^2-5x+6}{x^2-9}\le1$$

2. $$x^2-3x+2>0$$

3. $$\log(x^2-3x+2)\ne0$$

4. $$x^2-9\ne0$$

Now,

$$\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3},\qquad x\ne3$$

So,

$$-1\le\frac{x-2}{x+3}\le1$$

First solve

$$\frac{x-2}{x+3}\ge-1$$

$$\frac{x-2+x+3}{x+3}\ge0$$

$$\frac{2x+1}{x+3}\ge0$$

Hence,

$$x\in(-\infty,-3)\cup\left[-\frac12,\infty\right)$$

Now solve

$$\frac{x-2}{x+3}\le1$$

$$\frac{x-2-x-3}{x+3}\le0$$

$$\frac{-5}{x+3}\le0$$

$$x>-3$$

Combining both,

$$x\in\left[-\frac12,\infty\right)$$

Now,

$$x^2-3x+2=(x-1)(x-2)$$

For logarithm to exist,

$$(x-1)(x-2)>0$$

$$x<1\quad\text{or}\quad x>2$$

Also,

$$\log(x^2-3x+2)\ne0$$

$$x^2-3x+2\ne1$$

$$x^2-3x+1\ne0$$

This condition is automatically satisfied in the final interval.

Also,

$$x\ne3$$

Therefore, combining all conditions,

$$x\in\left[-\frac12,1\right)\cup(2,\infty)-\{3\}$$

Hence, the correct answer is

$$\boxed{\left[-\frac12,1\right)\cup(2,\infty)-\{3\}}$$