If $p \neq 0, q \neq 0$ and $$ \det\begin{bmatrix}p & q & p \alpha+q \\ q & r & q \alpha+r \\ p \alpha+q & q \alpha+r & 0\end{bmatrix}=0 $$ Then check
$\alpha$ is a root of the equation $p x^{2}+2 q x+r=0$ or not
My Approach:
I have put x = $\alpha$ in the equation $p \alpha^{2}+2 q \alpha+r=0$.....(1)
and I am getting the Determinant value as: $\begin{aligned}{p(q \alpha+r)^{2}+2\left(p q \alpha^{2}+q^{2} \alpha\right.}{+p r \alpha+2 r)} +(p \alpha+q)\left(q^{2} \alpha+p r \alpha\right) \end{aligned}$
So can I conclude the condition, $\alpha$ is a root of the equation $p x^{2}+2 q x+r=0$ is not true
Your determinant value seems to be wrong. The correct determinant is $$ -{\alpha}^{2}{p}^{2}r+{\alpha}^{2}p{q}^{2}-2\,\alpha pqr+2\,\alpha{q}^{3}-p{r}^{2}+{q}^{2}r =- \left( pr-{q}^{2} \right) \left( {\alpha}^{2}p+2\,q\alpha+r \right) $$ so unless $pr=q^2$, $\alpha$ is indeed a root of $p x^2 + 2 q x + r$.