For the system of equations
$$x + y + z = 6$$
$$x + 2y + \alpha z = 10$$
$$x + 3y + 5z = \beta$$, which one of the following is NOT true?

Given system: $$x + y + z = 6$$, $$x + 2y + \alpha z = 10$$, $$x + 3y + 5z = \beta$$.

Determinant of coefficient matrix:

$$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{vmatrix} = 1(10 - 3\alpha) - 1(5 - \alpha) + 1(3 - 2) = 6 - 2\alpha$$

For unique solution: $$D \neq 0 \implies \alpha \neq 3$$.

When $$\alpha = 3$$ ($$D = 0$$): Check consistency by row reduction.

$$R_2 - R_1$$: $$y + 2z = 4$$

$$R_3 - R_1$$: $$2y + 4z = \beta - 6$$

$$R_3 - 2R_2$$: $$0 = \beta - 6 - 8 = \beta - 14$$

If $$\beta = 14$$: infinitely many solutions. If $$\beta \neq 14$$: no solution.

Now checking each option:

Option A: $$\alpha = 3, \beta = 24$$. Since $$\beta \neq 14$$, no solution. TRUE. ✓

Option B: $$\alpha = -3, \beta = 14$$. $$D = 6 + 6 = 12 \neq 0$$, unique solution. TRUE. ✓

Option C: $$\alpha = 3, \beta = 14$$. Infinitely many solutions. TRUE. ✓

Option D: $$\alpha = 3, \beta \neq 14$$. Claims unique solution, but $$D = 0$$ when $$\alpha = 3$$, so the system has NO solution (not unique). FALSE. ✗

The statement that is NOT true is Option D.