For $$\alpha, \beta \in \mathbb{R}$$, suppose the system of linear equations
$$x - y + z = 5$$
$$2x + 2y + \alpha z = 8$$
$$3x - y + 4z = \beta$$
has infinitely many solutions. Then $$\alpha$$ and $$\beta$$ are the roots of

For the system to have infinitely many solutions, the determinant of the coefficient matrix must vanish; the equations are $$x - y + z = 5$$, $$2x + 2y + \alpha z = 8$$, and $$3x - y + 4z = \beta$$.

The determinant is given by $$D = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{vmatrix}$$. Expanding yields $$D = 1(8 + \alpha) - (-1)(8 - 3\alpha) + 1(-2 - 6) = 8 + \alpha + 8 - 3\alpha - 8 = 8 - 2\alpha$$, and setting $$D = 0$$ gives $$\alpha = 4$$.

With $$\alpha = 4$$, the first two equations become $$x - y + z = 5$$ and $$x + y + 2z = 4$$. Adding gives $$2x + 3z = 9 \Rightarrow x = \frac{9 - 3z}{2}$$, while subtracting gives $$-2y - z = 1 \Rightarrow y = \frac{-1 - z}{2}$$. Substituting into the third equation produces $$3\cdot\frac{9-3z}{2} - \frac{-1-z}{2} + 4z = \beta$$, which simplifies as $$\frac{27 - 9z + 1 + z}{2} + 4z = \frac{28 - 8z}{2} + 4z = 14 - 4z + 4z = 14$$, so $$\beta = 14$$.

Since the quadratic equation with roots $$\alpha = 4$$ and $$\beta = 14$$ has sum $$4 + 14 = 18$$ and product $$4 \times 14 = 56$$, it is given by $$x^2 - 18x + 56 = 0$$. Therefore the final answer is $$x^2 - 18x + 56 = 0$$.