If p, q, r are distinct, then value of […] is:

The correct answer is Option (1) → \((1+pqr)(q-p)(r-p)(r-q)\)

Each element of the third column is a sum, so the determinant splits into two determinants:

    \( \displaystyle \Delta = \begin{vmatrix}p & p^2 & 1\\ q & q^2 & 1\\ r & r^2 & 1\end{vmatrix} + \begin{vmatrix}p & p^2 & p^3\\ q & q^2 & q^3\\ r & r^2 & r^3\end{vmatrix} \)

In the second determinant, take p, q and r out of the three rows:

    \( \displaystyle \begin{vmatrix}p & p^2 & p^3\\ q & q^2 & q^3\\ r & r^2 & r^3\end{vmatrix} = pqr\begin{vmatrix}1 & p & p^2\\ 1 & q & q^2\\ 1 & r & r^2\end{vmatrix} \)

In the first determinant, bring the third column to the front by the interchanges \(C_3 \leftrightarrow C_2\), then \(C_2 \leftrightarrow C_1\). Two interchanges leave the sign unchanged:

    \( \displaystyle \begin{vmatrix}p & p^2 & 1\\ q & q^2 & 1\\ r & r^2 & 1\end{vmatrix} = \begin{vmatrix}1 & p & p^2\\ 1 & q & q^2\\ 1 & r & r^2\end{vmatrix} \)

So

    \( \displaystyle \Delta = (1+pqr)\begin{vmatrix}1 & p & p^2\\ 1 & q & q^2\\ 1 & r & r^2\end{vmatrix} \)

For this Vandermonde determinant, apply \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):

    \( \displaystyle \begin{vmatrix}1 & p & p^2\\ 0 & q-p & q^2-p^2\\ 0 & r-p & r^2-p^2\end{vmatrix} = (q-p)(r-p)\begin{vmatrix}1 & q+p\\ 1 & r+p\end{vmatrix} \)

    \( = (q-p)(r-p)(r-q) \)

Hence

    \( \Delta = (1+pqr)(q-p)(r-p)(r-q) \)