We have,
$$
\begin{aligned}
& \frac{x^2+5 x+7}{(x-3)^3}=\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{C}{(x-3)^3} \\
\Rightarrow x^2+5 x+7 & =A(x-3)^2+B(x-3)+C
\end{aligned}
$$
Put $x=3$ in Eq. (i), we get
$$
\begin{array}{rlrl}
9+15+7 & =C \\
\Rightarrow & & C & =31
\end{array}
$$
Put $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& 7=9 A-3 B+31 \\
& \Rightarrow \quad 9 A-3 B=-24 \\
& \Rightarrow \quad 3 A-B=-8 \\
&
\end{aligned}
$$
Put $x=1$ in Eq. (i), we get
$$
\begin{aligned}
& 1+5+7=4 A-2 B+31 \\
& \Rightarrow \quad 13=4 A-2 B+31 \\
& \Rightarrow \quad 4 A-2 B=-18 \\
& \Rightarrow \quad 2 A-B=-9 \\
& \text { On solving Eqs. (ii) and (iii), we get } \\
& A=1 \\
& \text { and } \\
& B=11 \\
&
\end{aligned}
$$
∴ Equation of line having slope A and pass ing
through the points (B, C) is
y − 31 = 1(x − 11)
y − 31 = x − 11
x − y + 20 = 0