Let the equation of plane passing through the line of intersection of the planes $$x + 2y + az = 2$$ and $$x - y + z = 3$$ be $$5x - 11y + bz = 6a - 1$$. For $$c \in \mathbb{Z}$$, if the distance of this plane from the point $$(a, -c, c)$$ is $$\frac{2}{\sqrt{a}}$$, then $$\frac{a+b}{c}$$ is equal to

We need to find $$\frac{a+b}{c}$$ given the conditions about the plane and the distance.

Find the plane through the intersection. The family of planes through the intersection of $$x + 2y + az = 2$$ and $$x - y + z = 3$$ is:

$$(x + 2y + az - 2) + \lambda(x - y + z - 3) = 0$$

$$(1+\lambda)x + (2-\lambda)y + (a+\lambda)z = 2 + 3\lambda$$

Compare with $$5x - 11y + bz = 6a - 1$$. The ratios must be equal:

$$\frac{1+\lambda}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$$

From the first two ratios:

$$-11(1+\lambda) = 5(2-\lambda)$$

$$-11 - 11\lambda = 10 - 5\lambda$$

$$-6\lambda = 21 \implies \lambda = -\frac{7}{2}$$

Find $$a$$ and $$b$$: $$1 + \lambda = -\frac{5}{2}$$, so the common ratio $$k = \frac{-5/2}{5} = -\frac{1}{2}$$

From $$\frac{2+3\lambda}{6a-1} = -\frac{1}{2}$$:

$$2 + 3\left(-\frac{7}{2}\right) = -\frac{1}{2}(6a-1) \implies -\frac{17}{2} = -3a + \frac{1}{2}$$

$$-3a = -9 \implies a = 3$$

$$b = \frac{a+\lambda}{-1/2} = \frac{3 - 7/2}{-1/2} = \frac{-1/2}{-1/2} = 1$$

Find the distance from $$(a, -c, c) = (3, -c, c)$$: $$\text{Distance} = \frac{|5(3) - 11(-c) + c - 17|}{\sqrt{25 + 121 + 1}} = \frac{|15 + 11c + c - 17|}{\sqrt{147}} = \frac{|12c - 2|}{\sqrt{147}}$$

This equals $$\frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}}$$:

$$\frac{|12c - 2|}{\sqrt{147}} = \frac{2}{\sqrt{3}} \implies |12c - 2| = \frac{2\sqrt{147}}{\sqrt{3}} = 2\sqrt{49} = 14$$

Solve for $$c$$: $$12c - 2 = 14 \implies c = \frac{16}{12} = \frac{4}{3}$$ (not an integer)

$$12c - 2 = -14 \implies c = \frac{-12}{12} = -1$$ ✓ (integer)

Calculate the answer: $$\frac{a+b}{c} = \frac{3+1}{-1} = -4$$

The correct answer is Option C: $$-4$$.