$\alpha ,\beta ,\gamma$ are roots of equation.
$\alpha +\beta +\gamma=0$
$\alpha \beta + \beta \gamma+ \gamma\alpha =-p$
$\alpha \cdot\beta \cdot\gamma=q$
The determinant of the matrix after simplification is
$3(\alpha \cdot\beta \cdot\gamma)-(\alpha^{3}+\beta^{3}+\gamma^{3})$
$3q-(\alpha^{3}+\beta^{3}+\gamma^{3}) $ ------------------$>(1)$
$\alpha^{3}+\beta^{3}+\gamma^{3}=(\alpha +\beta +\gamma)(\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta - \beta \gamma- \gamma\alpha )+3(\alpha \cdot\beta \cdot\gamma)=3q$
substitute the value of $\alpha^{3}+\beta^{3}+\gamma^{3}$ in equation $1$ and we get
$3q-3q=0$
So$,(C)$ is the correct choice.