If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of $$ \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{pmatrix} \,\,? $$ (A) $p^2+6q \quad$ (B) $1 \quad$ (C) $p \quad$ (D) $0 \quad$
In this equation given we have product of eigenvalues given as $-q$ and we know product of eigenvalues is determinant then why isn't the determinant is $-q$?
The statement you're thinking of is that the product of eigenvalues of a matrix $A$ is equal (up to sign, depending on parity) to the constant term of its characteristic polynomial $p_A$.
In this case, however, the given matrix, though constructed from the roots $a, b, c$ of the characteristic polynomial, doesn't have the property that its eigenvalues are $a, b, c$, and therefore the above general fact does not apply.
On the other hand, using, e.g., cofactor expansion gives that the determinant of the given matrix is $$\det \pmatrix{a&b&c\\b&c&a\\c&a&b} = 3 a b c - (a^3 + b^3 + c^3).$$ On the other hand, writing the cubic polynomial as $$x^3 - p x + q = (x - a)(x - b)(x - c),$$ expanding, and comparing like terms recovers a special case of Vieta's Formulas: $$\begin{array}{rcl} 0 &=& - (a + b + c) \\ -p &=& bc + ca + ab \\ q &=& -a b c \end{array} .$$ Manipulating these lets us write both terms of the expression $3 a b c - (a^3 + b^3 + c^3)$ for the determinant as polynomials in $p, q$. These manipulations give special cases of Newton's Identities.