If Δ=345x456y567zxyz0=0, then

Applying R1→R1+R3−2R2, we getΔ=000x+z−2y456y567zxyz0=−(x+z−2y)456567xyz                  [Expanding along R1]=−(x+z−2y)0−160−17x−2y+zy−zz      [Applying C1→C1+C3−2C2 and C2→C2−C3]=−(x+z−2y)2-16-17=(x−2y+z)2Hence, Δ=0⇒x, y, z are in A.P.