Given, $\begin{array}{r}\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2} \\ +\frac{C}{x-3}\end{array}$
$\begin{aligned} \Rightarrow x^2+x & +1=A(x-2)(x-3) \\ & +B(x-1)(x-3)+C(x-1)(x-2)\end{aligned}$
Put $x=1,2$ and 3 respectively we get
$\begin{array}{lc}
& 1+1+1=A(1-2)(1-3) \\
\Rightarrow & 3=2 A \Rightarrow A=\frac{3}{2} \\
& 2^2+2+1=B(2-1)(2-3) \\
\Rightarrow & 7=-B \Rightarrow B=-7 \\
\text { and } & 3^2+3+1=C(3-1)(3-2)
\end{array}$
$\begin{array}{ll}\Rightarrow & 13=2 C \Rightarrow C=\frac{13}{2} \\ \therefore & A+C=\frac{3}{2}+\frac{13}{2}=8\end{array}$