$\frac{3x^2+10x+13}{{(x-1)}^4}=\frac{A}{{(x-1)}^2}+\frac{B}{{(x-1)}^3}+\frac{C}{{(x-1)}^4}$ then descending order of  $A,B,C$ is

$\frac{3x^2+10x+13}{{(x-1)}^4}=\frac{A{(x-1)}^2+B(x-1)+C}{{(x-1)}^4}$

$3x^2+10x+13=A{(x-1)}^2+B(x-1)+C$

Put  $x=1\Rightarrow 3{\left(1\right)}^2+10\left(1\right)+13=A\left(0\right)+B\left(0\right)+C$

$3+10+13=C$

$C=26$

Compare the coefficient of $x^2\Rightarrow A=3$

Compare the constant term  $\Rightarrow A-B+C=13$

$\Rightarrow 3-B+26=13$

$\Rightarrow 29-B=13$

$\Rightarrow B=29-13$

$\Rightarrow B=16$

$\therefore$ Descending order  $C,B,A$.