If the shortest distance between the straight lines 3 ( x - 1 ) = 6 ( y - 2 ) = 2 ( z - 1 ) and 4 ( x - 2 ) = 2 ( y - λ ) = ( z - 3 ) , λ ∈ R is 1 38 , then the integral value of λ is equal to:

Step-by-step solution

The given lines are 3 ( x - 1 ) = 6 ( y - 2 ) = 2 ( z - 1 ) and 4 ( x - 2 ) = 2 ( y - λ ) = ( z - 3 ) And, they can be written as L 1 : ( x - 1 ) 2 = ( y - 2 ) 1 = ( z - 1 ) 3 and L 2 : ( x - 2 ) 1 = y - λ 2 = z - 3 4 . The lines are in the form x - x 1 l = y - y 1 m = z - z 1 n , where x 1 ,   y 1 ,   z 1 is a point on the line and l ,   m ,   n are the direction ratios of the vector parallel to the line. Thus, the position vectors of the points on the line are respectively a 1 &#8594