Answer:
Option A
Explanation:
△1=[1a2a31b2b31c2c3]
R2→R2−R1,R3→R3−R1
=[1a2a30b2−a2b3−a30c2−a2c3−a3]
=(b−a)(c−a)[1a2a30b+ab2+a2+ab0c2+a2c2+a+ac]
=(b−a)(c−a)[(b+a)(c2+a2+ac)−(c+a)(b2+a2+ab)]
=(b−a)(c−a)[bc2+a2b+abc+ac2+a3
+a2c−b2c−a2c−abc−b2a−a3−a2b]
=(b−a)(c−a)[(b−c)(ab+bc+ca)]
=−(a−b)(b−c)(c−a)(ab+bc+ca)
Now, △2=[bcb+c1cac+a1aba+b1]
R2→R2−R1 and R3→R3−R1
=[bcb+c1c(a−b)a−b0b(c−a)a−c0]=(a−b)(c−a)[bcb+c1c10a10]
=(a−b)(c−a)[1(c−b)]
=(a−b)(c−b)(c−a)
Now, △1△2= −(a−b)(b−c)(c−a)(ab+bc+ca)−(a−b)(b−c)(c−a)
△1△2 = (ab+bc+ca)