Given lines are $\frac{1-x}{3}$ = $\frac{y-2}{2\lambda}$ = $\frac{z-3}{2}$ and $\frac{x-1}{3\lambda}$ = $\frac{y-1}{1}$ = $\frac{6-z}{7}$ Let us rewrite the equations of the given lines as follows: $\frac{-(x-1)}{3}$ = $\frac{y-2}{2\lambda}$ = $\frac{z-3}{2}$ and $\frac{x-1}{3\lambda}$ = $\frac{y-1}{1}$ = $\frac{-(z-6)}{7}$ That is we have, $\frac{x-1}{-3}$ = $\frac{y-2}{2\lambda}$ = $\frac{z-3}{2}$ and $\frac{x-1}{3\lambda}$ = $\frac{y-1}{1}$ = $\frac{z-6}{-7}$ The lines are perpendicular so angle between them is 90 So, cos θ = 0 Here $(a_1,b_1,c_1)$ = (-3,2λ,2) and $(a_2,b_2,c_2)$ = (3λ,1,-7) For perpendicular lines $a_1a_2+b_1b_2+c_1c_2$ = 0 ⇒ - 9λ + 2λ - 14 = 0 ⇒ - 7λ = 14 ⇒ λ = $\frac{14}{-7}$ ⇒ λ = - 2