If the system of equations  2x+3y−z=0,x+ky−2z=3 and 2x−y+z=0  has a non-trivial solution  x,y,zthen xy+yz+zx+k  is equal to

Given system of linear equations 2x+3y−z=0x+ky−2z=0 And 2x−y+z=0 has a non-trivial solution (x,y,z)∴Δ=0⇒23−11k−22−11=02(k−2)−3(1+4)−1(1−2k)=0⇒2K−4−15+1+2K=0⇒ 4K=18⇒K=92So, system of linear equations is 2x+3y−z=0-----12x+9y−4z=0----2 And 2x−y+z=0-----3From Eqs. (i) and (ii) , we get 6y−3z=0,yz=12From Eqs. (i) and (iii), we get 4x+2y=0⇒xy=−12so,xz=xy×yz=−14⇒zx=−4                    ∵  yz=12   and  xy=−12∴ xy+yz+zx+k=−12+12−4+92=12