Let the lines ${\mathrm{l}}_1:\frac{\mathrm{x}+5}{3}=\frac{\mathrm{y}+4}{1}=\frac{\mathrm{z}-\mathrm{\alpha }}{-2}$ and ${\mathrm{l}}_2:3\mathrm{x}+2\mathrm{y}+\mathrm{z}-2=0=\mathrm{x}-3\mathrm{y}+2\mathrm{z}-13$ be coplanar. If the point P(a,b,c) on ${\mathrm{l}}_1$ is nearest to the point Q(-4,-3,2), then $\left|\mathrm{a}\right|+\left|\mathrm{b}\right|+\left|\mathrm{c}\right|$ is equal to

Plane passing through L₂ is of the form $(3\mathrm{x}+2\mathrm{y}+\mathrm{z}-2)+\mathrm{\lambda } (\mathrm{x}-3\mathrm{y}+2\mathrm{z}-13)=0$
Suppose L₁ lies on above plane then, $(3+\mathrm{\lambda })3+(2-3\mathrm{\lambda })\left(1\right)+(-2)(1+2\mathrm{\lambda })=0$
$\Rightarrow \mathrm{\lambda }=\frac{9}{4}$
$\therefore$ Plane containing L₁ & L₂ is $21\mathrm{x}-19\mathrm{y}+22\mathrm{z}-125=0$
$(-5,-4,\mathrm{\alpha })$  lies on the plane $\Rightarrow 22\mathrm{\alpha }=154\Rightarrow \mathrm{\alpha }=7$
DR’s of  PQ =  $(3\mathrm{\lambda }-1,\mathrm{\lambda }-1,5-2\mathrm{\lambda })$
$\therefore$$3(3\mathrm{\lambda }-1)+(\mathrm{\lambda }-1)+(-2)(5-2\mathrm{\lambda })=0$ 
$\Rightarrow \mathrm{\lambda }=1$
$\therefore$P= (-2,-3,5)=(a,b,c)
Now $\left|\mathrm{a}\right|+\left|\mathrm{b}\right|+\left|\mathrm{c}\right|=10$
   
 
  P= (-2,-3,5)=(a,b,c)
Now