If the system of linear equations, x + y + z = 6 x + 2 y + 3 z = 10 3 x + 2 y + λ z = μ has more than two solutions, then μ - λ 2 , is equal to.

Step-by-step solution

x + y + z = 6     . . . 1 x + 2 y + 3 z = 10     . . . 2 3 x + 2 y + λ z = μ     … 3 From 1   &   2 , If z = 0 ⇒ x + y = 6 and x + 2 y = 10 , ⇒ y = 4 ,   x = 2 If y = 0 ⇒ x + z = 6 and x + 3 z = 10 , ⇒ z = 2 ,   x = 4 So, 3 x + 2 y + λ z = μ , must pass through ( 2,4 , 0 ) and ( 4,0 , 2 ) . 6 + 8 = μ ⇒ μ = 14 And 12 + 2 λ = μ   ⇒ λ = 1 So, μ - λ 2 = 14 - 1 = 13