Please submit your concern

Explanation:

A = $\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$.

Determinant of A = a(cb - a²) + b(ac - b²) + c(ba - c²)

= 3abc - a³ - b³ - c³

= - (a³ + b³ + c³ - 3abc)

= - (a + b + c)(a² + b² + c² - ab - bc - ca)

(a + b + c)(a² + b² + c² - ab - bc - ca) > 0 for a, b, c > 0

∴ Determinant of A = - (a + b + c)(a² + b² + c² - ab - bc - ca) is always negative.

Hence, option (c).