x+ky-z=0 , 3x-ky-z=0 and x-3y+z=0 has non-zero solution for k=

If the system of equations x-ky-z=0 , kx-y-z=0 , x+y-z=0 has a non-zero solution , then possible values of k are :

The equations x+y-2z=0,2x-3y-z=0,x-5y+4z=k have non-zero solutions if k=

If x+y-z=0, 3x- alphay -3z=0, x-3y+z=0 has non zero solution , then alpha =

If the system of linear equations x+ky+3z=0 3x+ky-2z=0 2x+4y-3z=0 has a non-zero solution (x,y,z) then (xz)/(y^2) is equal to

If the system of linear equations x+ky+3z=0 3x+ky-2z=0 2x+4y-3z=0 has a non-zero solution (x,y,z) then (xz)/(y^2) is equal to

The given system of equations x-ky-z=0, kx-y-z=0, x+y-z=0 has non zero solution then the possible value of k is …..

The value of k for which the set of equations 3x+ky-2z=0,x+ky+3z=0 and 2x+3y-4z=0 has non-trivial solution is (A)15(B)16(C)31/2 (D) 33/2

If the system of equation x-(sin theta)y-z=0,(sin theta)x-y-z=0 and x+y-z=0 has non zero solution then (where pi in(0,pi))