If the lines $6x -2 = 3y + 1= 2z - 2$ and $\frac{x-2}{λ}=\frac{2y-5}{-3}, z = - 2$ are perpendicular, then λ = 

3

The equation of the given lines can be re-written as

$\frac{x-1/3}{1}=\frac{y+1/3}{2}=\frac{z-1}{3}$ and $\frac{x-2}{λ}=\frac{y-5/2}{-3/2}=\frac{z+2}{0}$

If these lines are perpendicular, then 

$1 × λ +2 × (-\frac{3}{2}) + 3 ×0=0 ⇒ λ-3 = 0 ⇒λ = 3$