Let Δ=1 a a2−bc1 b b2−ca1 c c2−ab , then∆ is equal to
0
a+b+c
12a2+b2+c2
none of these
Applying R2→R2−R1+R3→R3−R1 we get
Δ=(b−a)(c−a)1 a a2−bc0 1 a+b+c0 1 a+b+c=0
∵R2 and R3 are identical