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?

The given system of equations is:

$$ax + y + z = 1$$ ...(1)

$$x + ay + z = 1$$ ...(2)

$$x + y + az = \beta$$ ...(3)

The coefficient matrix is $$A = \begin{pmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{pmatrix}$$. The determinant of A is:

$$\det(A) = \begin{vmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{vmatrix}$$

Expanding along the first row:

$$\det(A) = a \begin{vmatrix} a & 1 \\ 1 & a \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & a \end{vmatrix} + 1 \begin{vmatrix} 1 & a \\ 1 & 1 \end{vmatrix}$$

$$= a(a^2 - 1) - 1(a - 1) + 1(1 - a)$$

$$= a^3 - a - a + 1 + 1 - a$$

$$= a^3 - 3a + 2$$

Factoring: $$a^3 - 3a + 2 = (a-1)^2(a+2)$$

Thus, $$\det(A) = (a-1)^2(a+2)$$

The system has a unique solution if $$\det(A) \neq 0$$, i.e., $$a \neq 1$$ and $$a \neq -2$$. Otherwise, we check consistency for $$a=1$$ or $$a=-2$$.

Now, evaluate each option:

Option A: It has infinitely many solutions if $$a = 2$$ and $$\beta = -1$$.

Substitute $$a=2$$, $$\beta=-1$$:

$$2x + y + z = 1$$ ...(i)

$$x + 2y + z = 1$$ ...(ii)

$$x + y + 2z = -1$$ ...(iii)

Since $$a=2 \neq 1,-2$$, $$\det(A) = (2-1)^2(2+2) = 1 \cdot 4 = 4 \neq 0$$, so unique solution exists. Subtract (i) and (ii):

$$(2x+y+z) - (x+2y+z) = 1-1 \implies x - y = 0 \implies x = y$$

Subtract (ii) and (iii):

$$(x+2y+z) - (x+y+2z) = 1 - (-1) \implies y - z = 2 \implies z = y - 2$$

Substitute into (i):

$$2y + y + (y-2) = 1 \implies 4y - 2 = 1 \implies 4y = 3 \implies y = \frac{3}{4}$$

Then $$x = \frac{3}{4}$$, $$z = \frac{3}{4} - 2 = -\frac{5}{4}$$. Unique solution: $$\left(\frac{3}{4}, \frac{3}{4}, -\frac{5}{4}\right)$$, not infinitely many. Thus, Option A is incorrect.

Option B: It has no solution if $$a = -2$$ and $$\beta = 1$$.

Substitute $$a=-2$$, $$\beta=1$$:

$$-2x + y + z = 1$$ ...(iv)

$$x - 2y + z = 1$$ ...(v)

$$x + y - 2z = 1$$ ...(vi)

Add (iv) and (v):

$$(-2x+y+z) + (x-2y+z) = 1+1 \implies -x - y + 2z = 2$$ ...(vii)

Multiply (vi) by -1:

$$-x - y + 2z = -1$$ ...(viii)

Equations (vii) and (viii) give $$2 = -1$$, a contradiction. Thus, no solution. Option B is correct.

Option C: $$x + y + z = \frac{3}{4}$$ if $$a = 2$$ and $$\beta = 1$$.

Substitute $$a=2$$, $$\beta=1$$:

$$2x + y + z = 1$$ ...(ix)

$$x + 2y + z = 1$$ ...(x)

$$x + y + 2z = 1$$ ...(xi)

Add (ix), (x), and (xi):

$$(2x+y+z) + (x+2y+z) + (x+y+2z) = 1+1+1 \implies 4x + 4y + 4z = 3 \implies x + y + z = \frac{3}{4}$$

Thus, Option C is correct.

Option D: It has infinitely many solutions if $$a = 1$$ and $$\beta = 1$$.

Substitute $$a=1$$, $$\beta=1$$:

$$x + y + z = 1$$ ...(xii)

$$x + y + z = 1$$ ...(xiii)

$$x + y + z = 1$$ ...(xiv)

All equations are identical, so infinitely many solutions (e.g., $$x = t$$, $$y = s$$, $$z = 1 - t - s$$ for any $$t,s$$). Thus, Option D is correct.

Options B, C, and D are correct, but Option A is not correct. Therefore, the statement that is NOT correct is Option A.