If a, b, c are three complex numbers such that a2+b2+c2=0 and Δ=b2+c2abacabc2+a2bcacbca2+b2=ka2b2c2, then the value of k is

Using a2+b2+c2=0 ,we can write ∆ as Δ=-a2abacab-b2bcacbc-c2=abc-aaab-bbcc-c [taking a, b, c common from C1, C2, C3 respectively]=a2b2c2-1111-1111-1 [taking a, b, c common from R1, R2, R3 respectively]=a2b2c202120100-1=-2a2b2c2210-1=4a2b2c2  [applying C1→C1+C3 and C3→C2+C3]=a2b2c202120100-1=-2a2b2c2210-1=4a2b2c2 [applying C1 → C1 + C3 and C3→ C2 + C3] Thus, k = 4.