The value of the determinant $\begin{vmatrix}bc&ca&ab\\p&q&r\\1&1&1\end{vmatrix},$ where a, b and c are respectively the pth, qth and rth terms of a H.P., is

0

The correct answer is option (1) : 0

Let D be the common difference of the corresponding A.P. and A be its first term. Then,

$\frac{1}{a}= A+(p-1)D, \frac{1}{b}=A+(q-1)D$ and $\frac{1}{c}= A+(r-1)D$

Now,

$\begin{vmatrix}bc & ca & ab\\p & q & r\\1 & 1 & 1\end{vmatrix}$

$=abc\begin{vmatrix}\frac{1}{a} & \frac{1}{b} & \frac{1}{c}\\p & q & r\\1 & 1 & 1\end{vmatrix}$     [Applying $R_1→R_1\left(\frac{1}{abc}\right)]$

$=abc \begin{vmatrix}\frac{1}{a} & \frac{1}{b} & \frac{1}{c}\\p-1 & q-1 & r-1\\1 & 1 & 1\end{vmatrix}$                [Applying $R_2→R_2-R-3$]

$=abc \begin{vmatrix}\frac{1}{a}-(p-1)D & \frac{1}{b}-(q-1)D & \frac{1}{c}-(r-1)D\\p-1 & q-1 & r-1\\1 & 1 & 1\end{vmatrix}$[Applying $R_1→R_1-DR_2$]

$=abc\begin{vmatrix}A& A& A\\p-1 & q-1 & r-1\\1 & 1 & 1\end{vmatrix}=abc×0=0$