For the system of linear equations
$$x + y + z = 6$$
$$\alpha x + \beta y + 7z = 3$$
$$x + 2y + 3z = 14$$
which of the following is NOT true?

Given system:

$$x + y + z = 6$$ ... (i)

$$\alpha x + \beta y + 7z = 3$$ ... (ii)

$$x + 2y + 3z = 14$$ ... (iii)

(iii) - (i): $$y + 2z = 8 \Rightarrow y = 8 - 2z$$ ... (iv)

From (i): $$x = 6 - y - z = 6 - (8-2z) - z = z - 2$$ ... (v)

$$\alpha(z-2) + \beta(8-2z) + 7z = 3$$

$$z(\alpha - 2\beta + 7) + (-2\alpha + 8\beta) = 3$$

For infinitely many solutions, this must be true for all $$z$$:

$$\alpha - 2\beta + 7 = 0$$ and $$-2\alpha + 8\beta = 3$$

From first: $$\alpha = 2\beta - 7$$. Substituting: $$-2(2\beta - 7) + 8\beta = 3 \Rightarrow -4\beta + 14 + 8\beta = 3 \Rightarrow 4\beta = -11 \Rightarrow \beta = -\frac{11}{4}$$

$$\alpha = 2(-\frac{11}{4}) - 7 = -\frac{11}{2} - 7 = -\frac{25}{2}$$

Check if $$(\alpha, \beta) = (-\frac{25}{2}, -\frac{11}{4})$$ is on $$x + 2y + 18 = 0$$:

$$-\frac{25}{2} + 2(-\frac{11}{4}) + 18 = -\frac{25}{2} - \frac{11}{2} + 18 = -18 + 18 = 0$$ ✓

Option (1): $$\alpha = \beta = 7$$, no solution

$$z(7 - 14 + 7) + (-14 + 56) = 0 + 42 = 42 \neq 3$$. Inconsistent, so no solution. ✓ TRUE.

Option (2): $$\alpha = \beta, \alpha \neq 7$$, unique solution

$$z(\alpha - 2\alpha + 7) + (-2\alpha + 8\alpha) = 3$$

$$z(7 - \alpha) + 6\alpha = 3$$

Since $$\alpha \neq 7$$, $$7 - \alpha \neq 0$$, so $$z = \frac{3 - 6\alpha}{7 - \alpha}$$, unique. ✓ TRUE.

Option (3): Unique $$(\alpha, \beta)$$ on $$x + 2y + 18 = 0$$ for infinitely many solutions

We found a unique such point $$(-25/2, -11/4)$$. ✓ TRUE.

Option (4): For every $$(\alpha, \beta) \neq (7,7)$$ on $$x - 2y + 7 = 0$$, infinitely many solutions

On $$\alpha - 2\beta + 7 = 0$$, so the coefficient of $$z$$ vanishes. Then $$-2\alpha + 8\beta = 3$$ must hold.

$$\alpha = 2\beta - 7$$, so $$-2(2\beta - 7) + 8\beta = -4\beta + 14 + 8\beta = 4\beta + 14 = 3 \Rightarrow \beta = -11/4$$.

So only $$\beta = -11/4$$ works, not "every point" on the line. For other points on $$x - 2y + 7 = 0$$, we get $$0 \cdot z + \text{(nonzero)} = 3$$... wait, we get $$-2\alpha + 8\beta \neq 3$$, so NO solution (not infinitely many).

So Option (4) is NOT true. ✓

The correct answer is Option (4): $$\boxed{$$ For every point $$(\alpha,\beta) \neq (7,7)$$ on $$x - 2y + 7 = 0,$$ the system has infinitely many solutions. $$}$$