L1:→r=(3+t)^i+(−1+2t)^j+(4+2t)^k
L1:x−31=y+12=z−42
⇒ D.R..S. of L1=1,2,2
L2:→r=(3+2s)^i+(3+2s)^j+(2+s)^k
L2:x−32=y−32=z−21
⇒ D.R.S. of L2=2,2,1
∴ D.R.S of required line L is parallel to D.R.S of (L1×L2)=(−2,3,−2)
∴ Equation of L:x2=y−3=z2
Solving L & L1
⇒(2λ,−3λ,2λ)=(μ+3,2μ−1,2μ+4)
⇒μ=−1,λ=1
So, intersection point P(2, -3, 2)
Let, Q(2v+3,2v+3,v+2)be required point on L2
Now, PQ=√(2v+3−2)2+(2v+3+3)2+(v+2−2)2=√17
⇒(2v+1)2+(2v+6)2+(v)2=17
⇒9v2+28v+20=0
⇒v=−2 (rejected), 109 (accepted)
∴Q=(3−209,3−209,2−109)
=(79,79,89)
∴18(a+b+c)
=18(79+79+89)
=44