The following system of linear equations
$$2x + 3y + 2z = 9$$
$$3x + 2y + 2z = 9$$
$$x - y + 4z = 8$$

We compute the determinant of the coefficient matrix: $$D = \begin{vmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{vmatrix}$$.

Expanding along the first row: $$D = 2(2 \cdot 4 - 2 \cdot (-1)) - 3(3 \cdot 4 - 2 \cdot 1) + 2(3 \cdot (-1) - 2 \cdot 1) = 2(8 + 2) - 3(12 - 2) + 2(-3 - 2) = 20 - 30 - 10 = -20$$.

Since $$D = -20 \neq 0$$, the system has a unique solution.

To find the solution, subtract the first equation from the second: $$(3x + 2y + 2z) - (2x + 3y + 2z) = 0$$, giving $$x - y = 0$$, so $$x = y$$.

Substituting $$x = y$$ into the third equation: $$x - x + 4z = 8$$, giving $$z = 2$$.

Substituting $$x = y$$ and $$z = 2$$ into the first equation: $$2x + 3x + 4 = 9$$, giving $$5x = 5$$, so $$x = 1$$.

The unique solution is $$(x, y, z) = (1, 1, 2)$$. We can verify: $$\alpha + \beta^2 + \gamma^3 = 1 + 1 + 8 = 10 \neq 12$$, so option (3) is incorrect. The system has a unique solution.