Let the lines $$L_1 : \frac{x+5}{3} = \frac{y+4}{1} = \frac{z-\alpha}{-2}$$ and $$L_2 : 3x + 2y + z - 2 = 0 = x - 3y + 2z - 13$$ be coplanar. If the point $$P(a, b, c)$$ on $$L_1$$ is nearest to the point $$Q(-4, -3, 2)$$, then $$|a| + |b| + |c|$$ is equal to

Given lines $$L_1: \frac{x+5}{3} = \frac{y+4}{1} = \frac{z-\alpha}{-2}$$ and $$L_2$$ is the intersection of planes $$3x + 2y + z = 2$$ and $$x - 3y + 2z = 13$$.

Find direction and point of $$L_2$$.

Direction of $$L_2$$ = $$\vec{n_1} \times \vec{n_2} = (3,2,1) \times (1,-3,2) = (7,-5,-11)$$

A point on $$L_2$$: Setting $$z = 0$$: $$3x + 2y = 2$$ and $$x - 3y = 13$$.

Solving: $$x = \frac{32}{11}$$, $$y = -\frac{37}{11}$$. Point $$B = \left(\frac{32}{11}, -\frac{37}{11}, 0\right)$$.

Find $$\alpha$$ using coplanarity condition.

Point on $$L_1$$: $$A = (-5, -4, \alpha)$$, direction $$\vec{d_1} = (3, 1, -2)$$.

$$\vec{d_1} \times \vec{d_2} = (3,1,-2) \times (7,-5,-11) = (-21, 19, -22)$$

$$\vec{AB} = \left(-\frac{87}{11}, -\frac{7}{11}, \alpha\right) \cdot (-21, 19, -22) = 0$$

$$\frac{87 \times 21}{11} - \frac{7 \times 19}{11} - 22\alpha = 0$$

$$\frac{1827 - 133}{11} = 22\alpha \implies \frac{1694}{11} = 22\alpha \implies \alpha = 7$$

Find point P on $$L_1$$ nearest to $$Q(-4, -3, 2)$$.

Parametric form of $$L_1$$: $$P = (-5+3t, -4+t, 7-2t)$$.

$$\vec{PQ} = (-4-(-5+3t), -3-(-4+t), 2-(7-2t)) = (1-3t, 1-t, -5+2t)$$

For nearest point, $$\vec{PQ} \cdot \vec{d_1} = 0$$:

$$3(1-3t) + 1(1-t) + (-2)(-5+2t) = 0$$

$$3 - 9t + 1 - t + 10 - 4t = 0 \implies 14 - 14t = 0 \implies t = 1$$

$$P = (-5+3, -4+1, 7-2) = (-2, -3, 5)$$

$$|a| + |b| + |c| = |-2| + |-3| + |5| = 2 + 3 + 5 = 10$$

Therefore, the correct answer is Option D: $$\mathbf{10}$$.