Let the system of linear equations
$$-x + 2y - 9z = 7$$
$$-x + 3y + 7z = 9$$
$$-2x + y + 5z = 8$$
$$-3x + y + 13z = \lambda$$
has a unique solution $$x = \alpha, y = \beta, z = \gamma$$. Then the distance of the point $$(\alpha, \beta, \gamma)$$ from the plane $$2x - 2y + z = \lambda$$ is

From the first three equations:

$$-x + 2y - 9z = 7$$ ... (1)

$$-x + 3y + 7z = 9$$ ... (2)

$$-2x + y + 5z = 8$$ ... (3)

(2)-(1): $$y + 16z = 2$$, so $$y = 2 - 16z$$ ... (4)

(3) - 2×(1): $$-3y + 23z = -6$$, so $$3y = 23z + 6$$ ... (5)

From (4) into (5): $$3(2-16z) = 23z + 6$$

$$6 - 48z = 23z + 6$$

$$71z = 0$$, $$z = 0$$

$$y = 2$$, from (1): $$-x + 4 = 7$$, $$x = -3$$

Check (4th eq): $$-3(-3) + 2 + 0 = 9 + 2 = 11$$, so $$\lambda = 11$$.

Point $$(\alpha, \beta, \gamma) = (-3, 2, 0)$$

Distance from plane $$2x - 2y + z = \lambda = 11$$, i.e., $$2x - 2y + z - 11 = 0$$:

$$d = \frac{|2(-3) - 2(2) + 0 - 11|}{\sqrt{4+4+1}} = \frac{|-6-4-11|}{3} = \frac{21}{3} = 7$$

This matches option 2: 7.