∣∣
∣∣x−6−12−3xx−3−32xx+2∣∣
∣∣=0
applying R1→R1−R2 gives
→(x−2)∣∣
∣∣13−12−3xx−3−32xx+2∣∣
∣∣=0
applying C3→C1+C3 gives
→(x−2)(x−1)∣∣
∣∣1302−3x1−32x1∣∣
∣∣=0
⇒(x−2)(x−1)(−5x−15)=0
⇒(x−1)(x−2)(x+3)=0
∴ Possible values of x are −3,1,2
Hence, sum of all values of x=−3+1+2=0