Given :
$\begin{vmatrix}xp+y&x&y\\yp+z&y&z\\0 &xp+y&yp+z\end{vmatrix}=0$
operating $C_1=C_1-PC_2-C_3$
$\begin{vmatrix}0&x&y\\0&y&z\\-(xp^2+2py+z) &xp+y&yp+z\end{vmatrix}=0$
$\Rightarrow (xz-y^2)(xp^2+2py+z)=0$
$\Rightarrow xz-y^2=0$
$\Rightarrow D=0$ for other quad.factor in p which shows second factor is a perfect square.
$\Rightarrow y^2=xz$
$\Rightarrow x,y,z$ are in G.P.
Hence (b) is the correct answer.