Let a, b, c ∈ R be all non-zero and satisfy a³ + b³ + c³ = 2. If the matrix A = [a, b, c; b, c, a; c, a, b] satisfies A?A = I, then a value of abc can be

A? A = I
⇒ a²+b²+c²=1 and ab+bc+ca=0
Now, (a+b+c)²=1 ⇒ a+b+c=±1
So, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca) = (±1) (1-0)=±1
⇒ 3abc = 2±1 = 3,1
⇒ abc = 1, 1/3