Given $\frac{2 x^2-3 x+5}{(x-7)^3}=\frac{A}{(x-7)}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$
$\begin{aligned}
\frac{2 x^2-3 x+5}{(x-7)^3} & =\frac{A(x-7)^2+B(x-7)+C}{(x-7)^3} \\
2 x^2-3 x+5 & =A\left(x^2+49-14 x\right)+B x-7 B+C \\
2 x^2-3 x+5 & =A x^2+x(B-14 A)+(49 A-7 B+C)
\end{aligned}$
Compare each coefficients.
$\begin{aligned}
& \Rightarrow \mathrm{A}=2, \\
& \Rightarrow \mathrm{B}-14 \mathrm{~A}=-3 \\
& \mathrm{~B}-14 \times 2=-3 \\
& \mathrm{~B}=-3+28=25 \\
& \Rightarrow 49 \mathrm{~A}-7 \mathrm{~B}+\mathrm{C}=5 \\
& 49 \times 2-7 \times 25+\mathrm{C}=5 \\
& 98-175+\mathrm{C}=5 \\
& \mathrm{C}-77=5 \\
& \mathrm{C}=77+5=82
\end{aligned}$
Now, $2 \mathrm{~A}-3 \mathrm{~B}+\mathrm{C}=2 \times 2-3 \times 25+82$
$=4-75+82=86-75=11$
Therefore, option (c) is correct.