Given $f(x)=2\ln(x-2)-x^2+ax+1$ and the largest open interval where $f$ is strictly increasing is $(2,3)$.

Find $a$ using $f'(x)$

For increasing, we need $f'(x)>0$.

$f'(x)=\frac{2}{x-2}-2x+a$

Simplify: $f'(x)=\frac{2+(a-2x)(x-2)}{x-2}$

Since domain is $x>2$, denominator is positive. So sign depends on numerator: $2+(a-2x)(x-2)$

Expand: $2+ax-2a-2x^2+4x = -2x^2+(a+4)x+(2-2a)$

We are told the largest interval of increase is $(2,3)$, so $f'(x)>0$ exactly for $2<x<3$, and at the endpoint $x=3$ we must have $f'(3)=0$

Now, $f'(3)=\frac{2}{1}-6+a=a-4$ Hence, $a=4$

Check: $f'(x)=\frac{2}{x-2}-2x+4$ $=\frac{2+(-2x+4)(x-2)}{x-2}$ Since $-2x+4=-2(x-2)$,

Now analyze $g(x)=(x-1)^3(x+2-a)^2$

Substitute $a=4$: $g(x)=(x-1)^3(x-2)^2$

Find interval where $g$ is strictly decreasing

Differentiate: $g'(x)=3(x-1)^2(x-2)^2+(x-1)^3\cdot 2(x-2)$

Factor: $g'(x)=(x-1)^2(x-2)\big[3(x-2)+2(x-1)\big]$ $g'(x)=(x-1)^2(x-2)(5x-8)$

Since $(x-1)^2\ge 0$, the sign of $g'(x)$ depends on $(x-2)(5x-8)$

Critical points are: $x=1,\quad x=\frac85,\quad x=2$

For decreasing, need $g'(x)<0$.

Check intervals:

• If $x<\frac85$, then $(x-2)<0$ and $(5x-8)<0$, so product $>0$
• If $\frac85<x<2$, then $(x-2)<0$ and $(5x-8)>0$, so product $<0$
• If $x>2$, then product $>0$

Therefore, the largest open interval where $g$ is strictly decreasing is $(b,c)=\left(\frac85,2\right)$

So, $b=\frac85,\qquad c=2$

Compute the required value $100(a+b-c)=100\left(4+\frac85-2\right)$ $=100\left(2+\frac85\right)=100\cdot \frac{18}{5}=360$

Final answer $100(a+b-c)=360$ So the correct option is A.