The correct option is D 4a2b2c2
Performing R1→R1−R2−R3
∣∣
∣
∣∣0−2c2−2b2b2c2+a2b2c2c2a2+b2∣∣
∣
∣∣
(−2)∣∣
∣
∣∣0c2b2b2c2+a2b2c2c2a2+b2∣∣
∣
∣∣
Performing R3→R3−R1
(−2)∣∣
∣
∣∣0c2b2b2c2+a2b2c20a2∣∣
∣
∣∣
Performing C3→C3−C1
(−2)∣∣
∣
∣∣0c2b2b2c2+a20c20a2−c2∣∣
∣
∣∣
(−2)[(−c2)(b2a2−b2c2)+b2(−c4−c2a2)]
=4a2b2c2