Let \((2,3)\) be the largest open interval in which the function \(f(x)=2 \log _{\mathrm{e}}(x-2)-x^2+a x+1\) is strictl…

\(\text{Given,}\\ \mathrm{f}\left(\mathrm{x}\right)=2{\log }_{\mathrm{e}}\left(\mathrm{x}-2\right)-{\mathrm{x}}^{2}+\mathrm{ax}+1\\ \text{On differentiating both sides, we get}\\ \mathrm{f}'\left(\mathrm{x}\right)=\frac{2}{\mathrm{x}-2}-2\mathrm{x}+\mathrm{a}\geq 0\left[\text{for increasing}f\left(x\right)\right]\\ \text{Again,differentiating both sides, we get}\\ \mathrm{f}"\left(\mathrm{x}\right)=\frac{-2}{(\mathrm{x}-2{)}^{2}}-2<0\\ f'\left(x\right)\text{ is decreasing function }\\ {\mathrm{f}}^{'}\left(3\right)\geq 0\\ 2-6+\mathrm{a}\geq 0\\ \mathrm{a}\geq 4\\ {\mathrm{a}}_{\min }=4\\ \text{Now, }\\ \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^{3}{\left(\mathrm{x}+2-\mathrm{a}\right)}^{2}\\ \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^{3}{\left(\mathrm{x}-2\right)}^{2}\\ \text{On differentiating, we get}\\ {\mathrm{g}}^{'}\left(\mathrm{x}\right)={\left(x-1\right)}^{3}2\left(x-2\right)+{\left(\mathrm{x}-2\right)}^{2}3{\left(x-1\right)}^{2}\\ ={\left(x-1\right)}^{2}\left(\mathrm{x}-2\right)\left(2\mathrm{x}-2+3\mathrm{x}-6\right)\\ ={\left(x-1\right)}^{2}\left(\mathrm{x}-2\right)\left(5\mathrm{x}-8\right)<0\\ \mathrm{x}\in \left(\frac{8}{5},2\right)\\ \text{Hence, }\\ 100\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right)=100\left(4+\frac{8}{5}-2\right)=360\)