The simultaneous equations kx+2y−z=1, (k−1)y−2z=2  and   (k+2)z=3  have only one solution when

The system of given equations are  kx+2y−z=1  .... ...(i)                                                        (k−1)y−2z=2  .... ...(ii)                                         and         (k+2)z=3   ... ...(iii)This system of equations has a unique solution, if  k2−10k−1−200k+2≠0⇒      (k+2)(k)(k−1)≠0⇒     k≠ −2,  0,  1ie,   k=−1,  is a required answer.