When ax²+bx+c is divided by (x+1), (x-2), then remainders are 2,7,respectively. (x-3) can divide it exactly. Then the value of c is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \begin{array}{l}\mathrm{f}(-1)=2\\ \mathrm{a}-\mathrm{b}+\mathrm{c}=2\to \left(1\right)\\ \mathrm{f}\left(2\right)=7\\ 4\mathrm{a}+2\mathrm{b}+\mathrm{c}=7\to \left(2\right)\\ \mathrm{f}\left(3\right)=0\\ 9\mathrm{a}+3\mathrm{b}+\mathrm{c}=0\to \left(3\right)\end{array} \end{gathered}$$

$$\begin{gathered} \left(1\right)-\left(2\right)\Rightarrow \mathrm{a}-\mathrm{b}+\mathrm{c}=2 \\ \frac{ -4\mathrm{a}\pm 2\mathrm{b}\pm \mathrm{c}=7}{-3\mathrm{a}-3\mathrm{b}=-5}\to \left(4\right) \end{gathered}$$

$$\begin{gathered} \left(2\right)-\left(3\right)\Rightarrow 4\mathrm{a}+2\mathrm{b}+\mathrm{c}=7 \\ \frac{ - 9\mathrm{a}\pm 3\mathrm{b}\pm \mathrm{c}=0}{-5\mathrm{a}-\mathrm{b}=7}\to \left(5\right) \end{gathered}$$

$$\begin{gathered} \left(4\right)-\left(5\right)\Rightarrow -3\mathrm{a}-3\mathrm{b}=-5 \\ \frac{ \mp 15\mathrm{a}\mp 3\mathrm{b}=-21}{ 12\mathrm{a}=-26} \end{gathered}$$

$\begin{array}{l}\mathrm{a}=-\frac{13}{6}\\ \mathrm{b}=\frac{23}{6},\mathrm{c}=8\end{array}$