Using properties of determinants, show that $p \alpha^2 + 2q \alpha + r = 0$ given that p, q and r are not in G.P and $ \begin{vmatrix} 1 & \dfrac{q}{p} & \alpha + \dfrac{q}{p} \\ 1 & \dfrac{r}{q} & \alpha + \dfrac{r}{q} \\ p \alpha + q & q \alpha + r & 0 \end{vmatrix} = 0 $
$R_1 \rightarrow R_1-R_2 \implies$
$
\begin{vmatrix}
0 & \dfrac{q}{p}-\dfrac{r}{q} & \dfrac{q}{p}-\dfrac{r}{q} \\
1 & \dfrac{r}{q} & \alpha+\dfrac{r}{q} \\
p\alpha + q & q \alpha + r & 0
\end{vmatrix} = 0
$
i.e. $\left(\dfrac{q}{p}-\dfrac{r}{q}\right)$
$
\begin{vmatrix}
0 & 1 & 1 \\
1 & \dfrac{r}{q} & \alpha+\dfrac{r}{q} \\
p\alpha+q & q\alpha+r & 0
\end{vmatrix} = 0
$
$C_2 \rightarrow C_2 - C_3 \implies \left(\dfrac{q^2-pr}{pq}\right)$
$
\begin{vmatrix}
0 & 0 & 1 \\
1 & -\alpha & \alpha+\dfrac{r}{q} \\
p\alpha+q & q\alpha+r & 0
\end{vmatrix} = 0
$
Expanding along $R_1$ gives
$\dfrac{\left(q^2-pr\right) \left[q\alpha+r +\alpha \left(p\alpha+q\right)\right]}{pq}= 0$
i.e. $\left(q^2-pr\right)\left(p\alpha^2 + 2q\alpha + r\right) = 0$ $\cdots$ (1)
Since p, q and r are not in G.P $\implies$ $q^2 \neq pr$
i.e. $q^2 - pr \neq 0$ $\cdots$ (2)
Therefore, from equations (1) and (2) we have
$p\alpha^2 + 2q \alpha + r = 0$