The values of k for which the system of equationsx+ky-3z=0 , 3x+ky-2z=0 , 2x+3y-4z=0 has a non-trivial solution is (are)
2110
3110
-5
4
From (1) and (2), 2x+z=0⇒2x=-z
From (3), we get 3y =5z
∴x=-12z,y=53z
Substituting this in (1) we get
-12z+5k3z-3z=0.⇒5k3-72z=0
Now , z≠0⇒k=2110