For the system of linear equations
$$2x - y + 3z = 5$$
$$3x + 2y - z = 7$$
$$4x + 5y + \alpha z = \beta$$,
which of the following is NOT correct?

The system of equations is $$2x - y + 3z = 5$$
$$3x + 2y - z = 7$$
$$4x + 5y + \alpha z = \beta$$.

The determinant of the coefficient matrix is $$D = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{vmatrix}$$. Expanding along the first row gives $$D = 2(2\alpha + 5) + 1(3\alpha + 4) + 3(15 - 8) = 4\alpha + 10 + 3\alpha + 4 + 21 = 7\alpha + 35 = 7(\alpha + 5)$$.

This determinant is zero exactly when $$\alpha = -5$$. For $$\alpha \neq -5$$ the determinant is nonzero, so the system has a unique solution in that case.

When $$\alpha = -5$$ and there are infinitely many solutions, the third equation must be a linear combination of the first two: $$(3) = p\,(1) + q\,(2)$$. Matching coefficients yields
$$2p + 3q = 4$$
$$-p + 2q = 5$$
$$3p - q = -5$$. Solving the second and third equations gives $$p = -1$$ and $$q = 2$$. Substituting these into the constant terms gives $$\beta = p(5) + q(7) = -5 + 14 = 9$$.

Now each option can be checked. Option (1) asserts infinitely many solutions when $$\alpha = -5$$ and $$\beta = 9$$, which agrees with the result above.

Option (2) proposes infinitely many solutions for $$\alpha = -6$$ and $$\beta = 9$$. However, when $$\alpha = -6$$ the determinant is $$D = 7(-6 + 5) = -7 \neq 0$$, so the system actually has a unique solution rather than infinitely many.

Option (3) states the system is inconsistent for $$\alpha = -5$$ and $$\beta = 8$$. In this case $$D = 0$$ but $$\beta \neq 9$$, so the third equation does not align with the first two and the system is indeed inconsistent.

Option (4) claims a unique solution for any $$\alpha \neq -5$$ (regardless of $$\beta$$), which is correct because the determinant is nonzero whenever $$\alpha \neq -5$$.

Therefore the statement that is not correct is Option (2).