Given that $a{\alpha }^2+2b\alpha +c\ne 0$and that the system of equations $(a\alpha +b)x+ay+bz=0,$$(b\alpha +c)x+by+cz=0$,$(a\alpha +b)y+(b\alpha +c)z=0$has a non trivial solution, then $a,b,c$are in 

$The condition for a homogeneous system of equations to have a non-trivial solution is \Delta =0$

$\Rightarrow \Delta =\left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ 0& a\alpha +b& b\alpha +c\end{array}\right|=0$

                $R_3\to R_3-(\alpha R_1+R_2)$

$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ -\left(a{\alpha }^2+2b\alpha +c\right)& 0& 0\end{array}\right|=0 \\ \Rightarrow -\left(a{\alpha }^2+2b\alpha +c\right)\left(ac-b^2\right)=0 \\ \Rightarrow ac-b^2=0 \left(\because a{\alpha }^2+2b\alpha +c\ne 0\right) \\ \Rightarrow b^2=ac \\ \Rightarrow a,b,c are in G.P. \end{gathered}$$