Let $$(2,3)$$  be the largest open interval in which the function $$f(x)=2\log_e(x-2)-x^2+ax+1$$ is strictly increasing and  $$(b,c)$$  be the largest open interval in which the function $$g(x)=(x-1)^3(x+2-a)^2$$ is strictly decreasing. Then  $$100(a+b-c)$$  is equal to:

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

Given $$f'(x) = 0$$ at $$x=3$$ (boundary of increasing interval):

$$\frac{2}{3-2} - 2(3) + a = 0 \implies 2 - 6 + a = 0 \implies \mathbf{a = 4}$$.

Analyze $$g(x)$$ with $$a=4$$.

$$g(x) = (x-1)^3(x-2)^2$$.

$$g'(x) = 3(x-1)^2(x-2)^2 + 2(x-1)^3(x-2) = (x-1)^2(x-2)[3(x-2) + 2(x-1)]$$

$$g'(x) = (x-1)^2(x-2)(5x - 8)$$.

For $$g(x)$$ to be strictly decreasing, $$g'(x) < 0$$.

Critical points are $$x=1, x=2, x=8/5$$. Since $$(x-1)^2$$ is always positive, we look at $$(x-2)(5x-8) < 0$$.

This happens when $$x \in (8/5, 2)$$. So, $$\mathbf{b = 1.6}$$ and $$\mathbf{c = 2}$$.

$$100(4 + 1.6 - 2) = 100(3.6) = 360$$.

Correct Option: B (360)