Let Δ=∣∣
∣∣1abc1bca1cab∣∣
∣∣.
Applying R2→R2−R1,R3→R3−R1, we get
Δ=∣∣
∣∣1abc0b−aca−bc0c−aab−bc∣∣
∣∣ =∣∣
∣
∣∣1abc0b−ac(a−b)0c−ab(a−c)∣∣
∣
∣∣
⇒Δ=(a−b)(a−c)∣∣
∣∣1abc0−1c0−1b∣∣
∣∣ [Taking (a-b) and (a-c)common from R1 and R3 respectively.]
⇒Δ=(a−b)(a−c)(−b+c) [Expanding along first column]
⇒Δ=(a−b)(b−c)(c−a)