The question is:
If $a,b,c$ are negative distinct real numbers,then the determinant $ \begin{vmatrix} a & b & c \\ b & c & a\\ c & a & b \end{vmatrix} $ is $(a) \le 0 \quad (b) \gt 0 \quad (c) \lt 0 \quad (d) \ge 0 $
My strategy: I identifed that the matrix is a circulant hence the determinant can be expressed in the form of $-(a^3 + b^3 + c^3 - 3abc)$ which implies that $-(a+b+c)\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$ hence $(-)(-)(+) \gt 0$ but the answers says it is $\ge 0$, so can we have three $a,b,c$ such that the answer is $0$ ?