The point of intersection $$C$$ of the plane $$8x + y + 2z = 0$$ and the line joining the points $$A(-3, -6, 1)$$ and $$B(2, 4, -3)$$ divides the line segment $$AB$$ internally in the ratio $$k : 1$$. If $$a, b, c$$ ($$|a|, |b|, |c|$$ are coprime) are the direction ratios of the perpendicular from the point $$C$$ on the line $$\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$$, then $$|a + b + c|$$ is equal to ______.

Find point $$C$$ where plane $$8x + y + 2z = 0$$ intersects line $$AB$$ with $$A(-3, -6, 1)$$ and $$B(2, 4, -3)$$.

Parametrize line $$AB$$.

$$P = A + t(B - A) = (-3 + 5t,\ -6 + 10t,\ 1 - 4t)$$

Find $$t$$.

$$8(-3 + 5t) + (-6 + 10t) + 2(1 - 4t) = 0$$

$$-24 + 40t - 6 + 10t + 2 - 8t = 0 \implies 42t = 28 \implies t = \frac{2}{3}$$

So $$C$$ divides $$AB$$ in ratio $$k:1 = 2:1$$.

$$C = \left(\frac{1}{3},\ \frac{2}{3},\ -\frac{5}{3}\right)$$

Perpendicular from $$C$$ to line $$\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$$.

Line point: $$(1, -4, -2)$$, direction: $$\vec{d} = (-1, 2, 3)$$.

$$\vec{AC} = C - (1, -4, -2) = \left(-\frac{2}{3},\ \frac{14}{3},\ \frac{1}{3}\right)$$

Project $$\vec{AC}$$ onto $$\vec{d}$$.

$$\vec{AC} \cdot \vec{d} = \frac{2}{3} + \frac{28}{3} + 1 = \frac{33}{3} = 11$$

$$|\vec{d}|^2 = 1 + 4 + 9 = 14$$

Perpendicular direction.

$$\vec{p} = \vec{AC} - \frac{11}{14}\vec{d} = \left(-\frac{2}{3} + \frac{11}{14},\ \frac{14}{3} - \frac{22}{14},\ \frac{1}{3} - \frac{33}{14}\right)$$

$$= \left(\frac{5}{42},\ \frac{130}{42},\ -\frac{85}{42}\right) = \frac{5}{42}(1, 26, -17)$$

Direction ratios: $$(a, b, c) = (1, 26, -17)$$ with $$|a|, |b|, |c|$$ pairwise coprime ✓

$$|a + b + c| = |1 + 26 - 17| = 10$$

The answer is $$\boxed{10}$$.