If […] =λ (x-y) (y-z) (z-x), then value of λ is :

The correct answer is Option (3) → $1+pxyz$

$\begin{vmatrix}x & x^2 & 1+px^3\\y & y^2 & 1+py^3\\z & z^2 & 1+pz^3\end{vmatrix}=\begin{vmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{vmatrix}+\begin{vmatrix}x & x^2 & px^3\\y & y^2 & py^3\\z & z^2 & pz^3\end{vmatrix}$

$=\begin{vmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{vmatrix}+pxyz\begin{vmatrix}1&x&x^2\\1&yy^2\\1&z&z^2\end{vmatrix}$

$=\begin{vmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{vmatrix}+pxyz\begin{vmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{vmatrix}$

$=(1-pxyz)\begin{vmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{vmatrix}$

$R_3→R_3-R_2,R_2→R_2-R_1$

$(1-pxyz)\begin{vmatrix}x & x^2 & 1\\y-x & y^2-x^2 & 0\\z-y & z^2-y^2 & 0\end{vmatrix}$

$=(1-pxyz)(z-y)(y-x)(y+z-x-y)$

$=(1-pxyz)(x-y)(y-z)(z-x)$

so $λ=1+pxyz$