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

We need to find which statement is NOT correct about the system: $$\alpha x + y + z = 1$$, $$x + \alpha y + z = 1$$, $$x + y + \alpha z = \beta$$.

First we write the coefficient matrix and find its determinant: $$D = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix}$$. Expanding gives $$D = \alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha)$$, which simplifies to $$\alpha^3 - \alpha - \alpha + 1 + 1 - \alpha$$, then to $$\alpha^3 - 3\alpha + 2$$, and finally to $$(\alpha - 1)^2(\alpha + 2)$$.

Option A: $$\alpha = 2, \beta = -1$$ -- infinitely many solutions. When $$\alpha = 2$$, the determinant becomes $$(2-1)^2(2+2) = 4 \neq 0$$, so the system has a unique solution. Thus the claim of infinitely many solutions is NOT correct.

Option B: $$\alpha = -2, \beta = 1$$ -- no solution. When $$\alpha = -2$$, the determinant is $$(-3)^2(0) = 0$$, so the system is singular. Adding all three equations: $$(-2+1+1)x + (1-2+1)y + (1+1-2)z = 1 + 1 + \beta$$ gives $$0 = 2 + \beta$$, and with $$\beta = 1$$ this leads to $$0 = 3$$, a contradiction. Therefore there is no solution, confirming the claim.

Option C: $$\alpha = 2, \beta = 1$$ -- $$x + y + z = 3/4$$. When $$\alpha = 2, \beta = 1$$ the determinant is $$4 \neq 0$$ so a unique solution exists. The system becomes $$2x + y + z = 1$$, $$x + 2y + z = 1$$, $$x + y + 2z = 1$$. Adding these yields $$4(x+y+z) = 3$$, giving $$x+y+z = 3/4$$, which verifies the claim.

Option D: $$\alpha = 1, \beta = 1$$ -- infinitely many solutions. When $$\alpha = 1$$, the determinant is $$0$$ and all three equations reduce to $$x + y + z = 1$$. With $$\beta = 1$$ all three are identical, leading to infinitely many solutions, so this claim is correct.

The statement that is NOT correct is Option A.