Solution:
\( L_1 : \dfrac{x+5}{3} = \dfrac{y+4}{1} = \dfrac{z-\alpha}{-2} \) pt.: \( (-5,-4,\alpha) \)
\( DR_1 = (3,1,-2) \)
\( L_2 : \) Intersection of planes \( 3x + 2y + z = 2, \ \ x - 3y + 2z = 13 \)

For pt. on \( L_2 \):
Let \( y = 0 \Rightarrow 3x + z = 2 \) and \( x + 2z = 13 \)
\( \Rightarrow \text{pt. } (x,y,z) = \left(\dfrac{-9}{5}, 0, \dfrac{37}{5}\right) \)
Now,
(Coplanar)
\( \overrightarrow{BA} \) vector is \( \perp \,r\) to \( (\overrightarrow{DR_1} \times \overrightarrow {DR_2}) \)

\( \overrightarrow{BA} = \dfrac{16}{5}\hat{i} + 4\hat{j} + \left(\dfrac{37}{5} - \alpha\right)\hat{k} \)
\( \Rightarrow \ (\overrightarrow {DR_1} \times \overrightarrow{DR_2})\cdot \overrightarrow{BA} = 0 \Rightarrow (-21)\times \dfrac{16}{5} + 4\times 19 + (-22)\left(\dfrac{37}{5} - \alpha\right) = 0 \)
on solving \( \alpha = 7 \).
Now in fig., any pt. on \( L_1 \) can be considered as:

\( (x,y,z) = ((3\lambda - 5), (\lambda - 4), (7 - 2\lambda)) \)
Nearest to \( (-4,-3,2) \)
\( \Rightarrow L_1 \) and formed line are \( \perp \ r. \)
\( \Rightarrow DR = (3\lambda - 1), (\lambda - 1), (5 - 2\lambda) \)
\( DR_1 = (3,1,-2) \)
\( \Rightarrow (3\lambda - 1)\cdot 3 + 1\cdot (\lambda - 1) + (-2)(5 - 2\lambda) = 0 \Rightarrow\quad \boxed{ \lambda = 1} \)
\( \Rightarrow \text{pt. } P(a,b,c) = ((3\cdot 1 - 5),(1 - 4),(7 - 2\cdot 1)) \)
\( p(a,b,c) = (-2,-3,5) \)
\( \Rightarrow |a| + |b| + |c| = |-2| + |-3| + |5| = 10 \)

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