let \(\mathrm{f}:\mathrm{R}\to \mathrm{R}\) be a function defined by \(f(x)=(2+3a){x}^{2}+\left(\frac{a+2}{a-1}\right)x+…

\(\text{Given, }\\ f(x)=(3a+2){x}^{2}+\left(\frac{a+2}{a-1}\right)x+b\\ f(x+y)=f(x)+f(y)+1-\frac{2}{7}xy...\left(i\right)\\ \text{put y}=\frac{1}{2},\text{we get}\\ f(x+\frac{1}{2})=f(x)+f(\frac{1}{2})+1-\frac{2}{7}x\frac{1}{2}\\ f(x+\frac{1}{2})=f(x)+f(\frac{1}{2})+1-\frac{1}{7}x...(ii)\\ \text{Now, put x=y=0}\\ \text{then f}\left(0\right)=2f\left(0\right)+1\\ f\left(0\right)=-1\\ \text{ So, }f(0)=0+0+b=-1\\ \Rightarrow b=-1\\ \text{ In (1) Put }y=-x\\ \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7}{x}^{2}\\ -1=2(3a+2){x}^{2}+2b+1+\frac{2}{7}{x}^{2}\\ -1=\left(2(3a+2)+\frac{2}{7}\right){x}^{2}+1-2\\ \Rightarrow 6a+4+\frac{2}{7}=0\\ a=-\frac{5}{7}\\ \text{put the value of a and b, we get}\\ \text{ So }f(x)=-\frac{1}{7}{x}^{2}-\frac{3}{4}x-1\\ \Rightarrow |\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4{\mathrm{x}}^{2}+21\mathrm{x}+28\right|\\ \text{ Now, }\\ 28\sum _{i=1}^{5}|f(i)|=28(\left|f\left(1\right)+f(2)+\ldots +f(5)\right|)\\ 28\sum _{i=1}^{5}|f(i)|=28\times \frac{1}{28}\times 675=675\)