The correct option is C (a+b+c)(a2+b2+c2)
∣∣
∣∣ab−cc+ba+cbc−aa−bb+ac∣∣
∣∣
C1→aC1
=1a∣∣
∣
∣∣a2b−cc+ba2+acbc−aa2−abb+ac∣∣
∣
∣∣
C1→C1+bC2+cC3
=1a∣∣
∣
∣∣a2+b2+c2b−cc+ba2+b2+c2bc−aa2+b2+c2b+ac∣∣
∣
∣∣
=a2+b2+c2a∣∣
∣∣1b−cc+b1bc−a1b+ac∣∣
∣∣
R2→R2−R1,R3→R3−R1
=a2+b2+c2a∣∣
∣∣1b−cc+b0c−(a+b)0a+c−b∣∣
∣∣
=a2+b2+c2a[−bc+a2+ac+bc+ab]
=(a+b+c)(a2+b2+c2)