[Solution by Dr. Abubacker] $f'(x)=\frac2{x-2}-2x+a$ must vanish at $x=3$: $2-6+a=0\Rightarrow a=4$, and then $f'=\frac{-2(x-1)(x-3)}{x-2}>0$ on $(2,3)$. For $g(x)=(x-1)^3(x-2)^2$: $g'=(x-1)^2(x-2)(5x-8)<0$ for $\frac85<x<2$. So $b=\frac85$, $c=2$ and $100(4+1.6-2)=360$. Answer (2).