Let Δ=∣∣−a2abacab−b2bcacbc−c2∣∣
Taking a,b,c common from R1,R2 and R3 respectively, we get. Δ=abc∣∣−aaab−bbcc−c∣∣]=a2b2c2[∣∣−1111−1111−1∣∣
taking a, b, c common from C1,C2,C3 respectively] =a2b2c2∣∣−111002020∣∣
( applying C2→C2+C1,C3→C3+C1) =a2b2c2⋅(−1)∣∣0220∣∣∣=a2b2c2(−1)(0−4) ⇒Δ=4a2b2c2