If the system of linear equations x + k y + 3 z = 0 3 x + k y - 2 z = 0 2 x + 4 y - 3 z = 0 has a non-zero solution x , y , z , then x z y 2 is equal to:

Step-by-step solution

1 k 3 3 k - 2 2 4 - 3 = 0 ⇒ - 3 k + 8 - 3 - 3 k - 12 + 2 - 5 k = 0 ⇒ - 4 k + 44 = 0 ⇒ k = 11 ⇒ x + 11 y + 3 z = 0       . . . 1 ⇒ 3 x + 11 y - 2 z = 0       . . . 2 ⇒ 2 x + 4 y - 3 z = 0       . . . 3 On solving equations 1   &   2 , we get 2 x - 5 z = 0 ⇒ x = 5 z 2 Put it in equation 3 , we get 2 z + 4 y = 0 ⇒ z = - 2 y ∴ x z y 2 = - 2 y × - 5 y y 2 = 10