R1 = R1 + R2 + R3
thus row 1 becomes
2(a+b+c), 2(a+b+c), 2(a+b+c)
we can take 2(a+b+c) outside
R1 is 1,1,1
NOw do C1 = C1 - C3 and C2 = C2 - C3
C1 = 0, a -b, b - c
C2 = 0, a -c, b -a
C3 = 1,b+c,c+a
Thus determinant becomes
2(a+b+c)( 1 ( (a-b)(b-a) - ( b-c)(a-c ) ) = 0
2(a+b+c)( - (a^2 + b^2 - 2ab ) - (ab -ac - bc + c^2 ) )
2(a+b+c)( -(a^2 + b^2 +c^2 -ab -bc -ca ) = 0
-2(a+b+c)(a^2 + b^2 +c^2 -ab -bc -ca ) = 0
thus either
a+b+c = 0
or
a^2 + b^2 +c^2 -ab -bc -ca = 0
2(a^2 + b^2 +c^2 -ab -bc -ca) = 0
(a-b)^2 + (b-c)^2 + (c-a)^2 = 0
which is possible only if a=b=c
