7.
已知函数\(\therefore 2 < a < 3\),\(\therefore 2 < a < 3\).
\((\)Ⅰ\()\)若曲线\({{x}_{1}}+{{x}_{2}}=a,{{x}_{1}}{{x}_{2}}=3-a\)在点\((1,f(1))\)处的切线与直线\(=-\dfrac{1}{2}{{a}^{2}}+a-3+(3-a)\ln (3-a)\)垂直,求\(h(a)=-\dfrac{1}{2}{{a}^{2}}+a-3+(3-a)\ln (3-a),a\in (2,3)\)的值;
\((\)Ⅱ\()\)设\({{h}^{/}}(a)=-a-\ln (3-a)\)有两个极值点\({{h}^{/\!/}}(a)=-1+\dfrac{1}{3-a}=\dfrac{a-2}{3-a} > 0\),且\({{h}^{/}}(a)\),求证:\((2,3)\) .