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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,

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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: $0 - \alpha(a \alpha+b) - (b \alpha+c) = -(a \alpha^2+2 b \alpha+c)$, $a \alpha+b - \alpha(a) - b = 0$, and $b \alpha+c - \alpha(b) - c = 0$.Thus, the determinant becomes:$\left|\begin{array}{ccc}a \alpha+b & a & b \ b \alpha+c & b & c \ -(a \alpha^2+2 b \alpha+c) & 0 & 0\end{array}ight|=0$Expanding along the third row:$-(a \alpha^2+2 b \alpha+c)(ac - b^2) = 0$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 $a, b, c$ are in Geometric Progression.

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