If x → and y → are two non zero, non-collinear vectors satisfying a - 3 α 2 + b - 4 α + c - 1 x → + a - 3 β 2 + b - 4 β + c - 1 y → + a - 3 γ 2 + b - 4 γ + c - 1 x → × y → = 0 (where α , β , γ are three distinct numbers), then the value of a 2 + b 2 + c 2 4 is equal to

Step-by-step solution

Since, x → , y → and x → × y → are 3 non-coplanar vectors Hence, a - 3 α 2 + b - 4 α + c - 1 = 0 a - 3 β 2 + b - 4 β + c - 1 = 0 a - 3 γ 2 + b - 4 γ + c - 1 = 0 We can say that a - 3 x 2 + b - 4 x + c - 1 = 0 has roots α , β , γ Since, quadratic has 3 roots so it becomes an identity ⇒ a - 3 = 0 , b - 4 = 0 , c - 1 = 0 ⇒ a = 3 , b = 4 , c = 1 ⇒ a 2 + b 2 + c 2 4 = 26 4 = 6 . 5