For the system of linear equations $\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta$, which one of the following statements is NOT correct?

For infinite solution $\Delta=\Delta_x=\Delta_y=\Delta_z=0$ <br/><br/>$$ \Delta=\left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2 $$ <br/><br/>If $\beta=1$, then all planes are overlapping <br/><br/>$\therefore$ Option (A) is correct. <br/><br/><b>Option (B) :</b> <br/><br/>If $\alpha=2 \Rightarrow \Delta \neq 0$ <br/><br/>$\therefore $ Unique solution exist <br/><br/>$\therefore$ Option (B) is incorrect. <br/><br/><b>Option (C) :</b> <br/><br/>$$ \begin{aligned} & \alpha=2, \beta=1 \\\\ & 2 x+y+z=1 \\\\ & x+2 y+z=1 \\\\ & x+y+2 z=1 \end{aligned} $$ <br/><br/>Adding all three equations, <br/><br/>$x+y+z=\frac{3}{4}$ <br/><br/>$\therefore$ option (C) is correct. <br/><br/><b>Option (D) :</b> <br/><br/>If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$ <br/><br/>$\therefore$ No solution. <br/><br/>$\therefore$ Option (D) is correct.