For lines to be perpendicular, we have
a1a2 + b1b2 + c1c2 = 0
⇒ -3 x 3λ + 2λ + 2 + 2 x -5 = 0
⇒ -9λ + 4λ - 10 = 0
⇒ -5λ + 4λ - 10 = 0
⇒ -5λ = 10
⇒ λ = 10/-5 = -2
∴ Given lines are
\(\frac{x-1}{-3} = \frac{y-2}{-4}=\frac{z-3}2=s\)
and \(\frac{x-1}{-6} = \frac{y-2}{2}=\frac{z-2}{-5}=t\)
⇒ x = 1 - 3s, y = 2 - 4s, z = 3 + 2s
∴ (1 - 3s, 2 - 4s, 3 + 2s) will lie on line (1)
& x = 1 - 6t, y = 1 + 2t, z = 6 - 5t
∴ (1 - 6t, 1 + 2t, 6 - 5t) will lie on line (2)
If both lines intersects, then
1 - 3s = 1 - 6t
⇒ 3s - 6t = 0 ⇒ s = 2t
& 2 - 4s = 1 + 2t ⇒ 4s = 1 - 2t ⇒ 8t = 1 - 2t
⇒ t = 1/10
∴ s = 2/10 = 1/5
and 3 + 2s = 6 - 5t
⇒ 3 + 2/5 = 6 - 5/10
⇒ 17/5 = 11/2 (Not satisfies)
∴ Both lines did not intersect each other.