Let the line $$\frac{x}{1} = \frac{6-y}{2} = \frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1}$$ and $$\frac{x+3}{6} = \frac{3-y}{3} = \frac{z-6}{1}$$ at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane $$2x - 2y + z = 14$$ is

Line 1: $$\frac{x}{1} = \frac{6-y}{2} = \frac{z+8}{5} = t$$. Point: $$(t, 6-2t, 5t-8)$$.

Line 2: $$\frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1} = s$$. Point: $$(5+4s, 7+3s, -2+s)$$.

At A: $$t = 5+4s, 6-2t = 7+3s, 5t-8 = -2+s$$.

From eq 1&2: $$t = 5+4s$$ and $$6-2(5+4s) = 7+3s$$, $$-4-8s = 7+3s$$, $$s = -1$$, $$t = 1$$.

A = (1, 4, -3). Check: 5(1)-8 = -3, -2+(-1) = -3 ✓

Line 3: $$\frac{x+3}{6} = \frac{3-y}{3} = \frac{z-6}{1} = u$$. Point: $$(-3+6u, 3-3u, 6+u)$$.

At B: $$t = -3+6u, 6-2t = 3-3u, 5t-8 = 6+u$$.

From eq 1&2: $$t = -3+6u$$ and $$6-2(-3+6u) = 3-3u$$, $$12-12u = 3-3u$$, $$9u = 9$$, $$u = 1$$, $$t = 3$$.

B = (3, 0, 7). Check: 5(3)-8 = 7, 6+1 = 7 ✓

Midpoint M = (2, 2, 2). Distance from $$2x-2y+z=14$$:

$$\frac{|4-4+2-14|}{3} = \frac{12}{3} = 4$$.

The correct answer is Option 3: 4.