Given that,a alpha^2+2 b alpha+c neq 0 and th | vedclass

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(B) The given system of equations is:$(a \alpha+b) x+a y+b z = 0$$(b \alpha+c) x+b y+c z = 0$$(a \alpha+b) y+(b \alpha+c) z = 0$For a non-trivial solu

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Geometric progression

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(B) The given system of equations is:$(a \alpha+b) x+a y+b z = 0$$(b \alpha+c) x+b y+c z = 0$$(a \alpha+b) y+(b \alpha+c) z = 0$For a non-trivial solution,the determinant of the coefficient matrix must be zero:$\left|\begin{array}{ccc}a \alpha+b & a & b \ b \alpha+c & b & c \ 0 & a \alpha+b & b \alpha+c\end{array}ight|=0$Applying the row operation $R_3 ightarrow R_3 - \alpha R_1 - R_2$:The third row becomes: $0 - \alpha(a \alpha + b) - (b \alpha + c) = -a \alpha^2 - b \alpha - b \alpha - c = -(a \alpha^2 + 2b \alpha + c)$.The determinant simplifies to:$\left|\begin{array}{ccc}a \alpha+b & a & b \ b \alpha+c & b & c \ -(a \alpha^2+2 b \alpha+c) & 0 & 0\end{array}ight|=0$Expanding along the third row:$-(a \alpha^2+2 b \alpha+c) 	imes (ac - b^2) = 0$Since it is given that $a \alpha^2+2 b \alpha+c 
eq 0$,we must have:$ac - b^2 = 0 \Rightarrow b^2 = ac$This condition implies that $a, b,$ and $c$ are in Geometric Progression $(GP)$.

Given that,$a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution,then $a, b$ and $c$ lie in

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Given that,$a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$has a non-trivial solution,then $a, b$ and $c$ lie in