Given:

$\frac{x^2-3x+2}{\left(x-4\right){\left(x-3\right)}^2}=\frac{A}{\left(x-4\right)}+\frac{B}{\left(x-3\right)}+\frac{C}{{\left(x-3\right)}^2}$ $\Rightarrow \frac{x^2-3x+2}{\left(x-4\right){\left(x-3\right)}^2}=\frac{A{\left(x-3\right)}^2+B\left(x-3\right)\left(x-4\right)+\left(x-4\right)}{\left(x-4\right){\left(x-3\right)}^2}$
$\Rightarrow x^2-3x+2=A\left(x^2+9-6x\right)+B\left(x^2-7x+12\right)+(Cx-4C)$

$\Rightarrow x^2-3x+12=(A+B)x^2+(-6A-7B+C)x+9A+12B-4C$

So,

$A+B=1 ...\left(1\right)$

$-6A-7B+C=-3 ...\left(2\right)$

$9A+12B-4C=2 ...\left(3\right)$

On solving, we get

$A=6, B=-5, C=-2$

So, $A+B+C=-1$