The curve $y=x^3-3x+2andx+3y=2$  intersect in points $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$  then the point P (A , B) where $A={\displaystyle \sum x_{i}andB=}{\displaystyle \sum y_i}$ lies on the line

Eliminating y we get $x^3-3x+2=\frac{2-x}{3}$                                                                                                                                                                                                           $3x^3-8x+4=0$                                                                                                                           $\Rightarrow {\displaystyle \sum x_i}=0{\displaystyle \sum y_i}={\displaystyle \sum _{i=1}^3\left(\frac{2}{3}-\frac{1}{3}{x}_i\right)}=\frac{6}{3}=2$                                                                                  So, coordinate of P are (0 , 2) only satisfies 4th option