For what values of k, the system of linear equations x+y+z=2, 2x+y-z=3, 3x+2y+kz=4 has a unique solution?

From x+y+z=2, we get x=2-y-z, substitute this in remaining two equations, we get 2(2-y-z)+y-z=3, which implies y+3z=1. 1) 3(2-y-z)+2y+kz=4, which implies y+(3-k)z=2. 2) Subtract equation 2) from 1), we get kz=-1, which gives z=-1/k. Therefore for k≠0, the system of equations have a unique solution.