Consider the following system of equations
$$\alpha x + 2y + z = 1$$
$$2\alpha x + 3y + z = 1$$
$$3x + \alpha y + 2z = \beta$$
For some $$\alpha, \beta \in \mathbb{R}$$. Then which of the following is NOT correct.

To determine which statement is not correct, we need to analyze the determinant of the coefficient matrix and the conditions for consistency.

The given system is:

$$\alpha x + 2y + z = 1$$

$$2\alpha x + 3y + z = 1$$

$$3x + \alpha y + 2z = \beta$$

Let A be the coefficient matrix of the system:

$$A = \begin{pmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{pmatrix}$$

First, calculate the determinant of the coefficient matrix A.

$$|A| = \alpha(3(2) - 1(\alpha)) - 2(2\alpha(2) - 1(3)) + 1(2\alpha(\alpha) - 3(3))$$

$$|A| = \alpha(6 - \alpha) - 2(4\alpha - 3) + (2\alpha^2 - 9)$$

$$|A| = 6\alpha - \alpha^2 - 8\alpha + 6 + 2\alpha^2 - 9$$

$$|A| = \alpha^2 - 2\alpha - 3$$

$$|A| = (\alpha - 3)(\alpha + 1)$$

The system will not have a unique solution when the determinant is equal to zero, which happens when $$\alpha = 3$$ or $$\alpha = -1$$.

Next, let us check the case when $$\alpha = 3$$.

Substitute this into the original equations:

$$3x + 2y + z = 1$$ (Equation 1)

$$6x + 3y + z = 1$$ (Equation 2)

$$3x + 3y + 2z = \beta$$ (Equation 3)

Subtracting 2 times Equation 1 from Equation 2 gives:

$$-y - z = -1 \implies y + z = 1$$

Subtracting Equation 1 from Equation 3 gives:

$$y + z = \beta - 1$$

For the system to be consistent, the right sides must match:

$$1 = \beta - 1 \implies \beta = 2$$

Thus, if $$\alpha = 3$$, the system has infinitely many solutions if $$ \beta = 2 $$, and no solution if $$\beta \neq 2$$. 

Now, let us check the case when $$\alpha = -1$$.

Substitute this into the original equations:

$$-x + 2y + z = 1$$ (Equation 1)

$$-2x + 3y + z = 1$$ (Equation 2)

$$3x - y + 2z = \beta$$ (Equation 3)

Subtracting 2 times Equation 1 from Equation 2 gives:

$$-y - z = -1 \implies y + z = 1$$

Multiplying Equation 1 by 3 and adding it to Equation 3 gives:

$$5y + 5z = \beta + 3 \implies 5(y + z) = \beta + 3$$

Substituting $$y + z = 1$$ into this equation yields:

$$5(1) = \beta + 3 \implies \beta = 2$$

Thus, if $$ \alpha = -1 $$, the system has infinitely many solutions if $$ \beta = 2 $$, and no solution if $$\beta \neq 2$$. 

Therefore, the statement that is NOT correct is:

It has no solution for $$\alpha = -1$$ and for all $$\beta \in \mathbb{R}$$