Answer 1

`("x"-5)/(5λ+2) = ("y"-2)/(-5) = ("z"-1)/(1)`                      ....(1)

and

`("x")/(1) = ("y"+1/2)/(2λ) = ("z" -1)/(3)`                              ...(2)

a₁ = 5λ + 2, b₁ = -5, c₁ = 1 and

a₂ = 1, b₂ = 2λ, c₂ = 3

a₁a₂ + b₁b₂ + c₁c₂ = 0

(5λ + 2) - 5(2λ) + 1(3) = 0

-5λ + 5 = 0

⇒ λ = -1