Consider the following system of questions α x+2y+z=1 2α x+3y+z=1 3x+α y+2z=β For some α ,β ∈ mathbbR. Then which of the following is NOT correct.

Given,

System of questions

$$\alpha x+2y+z=1$$αx+2y+z=1

$$2\alpha x+3y+z=1$$2αx+3y+z=1

$$3x+\alpha y+2z=\beta$$3x+αy+2z=β

Now finding, $D=|\begin{matrix}\alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2\end{matrix}|=0$D=∣α2α3​23α​112​∣=0

$$\Rightarrow \alpha (6-\alpha )-2(4\alpha -3)+1(2{\alpha }^{2}-9)=0$$⇒α(6−α)−2(4α−3)+1(2α2−9)=0

$$\Rightarrow (\alpha -3)(\alpha +1)=0$$⇒(α−3)(α+1)=0

$$\Rightarrow \alpha =-1,3$$⇒α=−1,3

$${D}_{x}=|\begin{matrix}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta \end{matrix}|=0\Rightarrow \beta =2$$Dx​=∣23α​112​11β​∣=0⇒β=2

$$\Rightarrow 2(\beta -2)-1(3\beta -\alpha )+(6-\alpha )=0$$⇒2(β−2)−1(3β−α)+(6−α)=0

$$\Rightarrow \beta =2$$⇒β=2

$${D}_{y}=|\begin{matrix}\alpha & 1 & 1 \\ 2\alpha & 1 & 1 \\ 3 & 2 & \beta \end{matrix}|=0$$Dy​=∣α2α3​112​11β​∣=0

$$\Rightarrow \alpha (\beta -2)-1(2\alpha \beta -3)+1(4\alpha -3)=0$$⇒α(β−2)−1(2αβ−3)+1(4α−3)=0

$$\Rightarrow \alpha (2-\beta )=0$$⇒α(2−β)=0

$$\Rightarrow \alpha =0,\beta =2$$⇒α=0,β=2

$${D}_{z}=|\begin{matrix}\alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & \beta \end{matrix}|=0$$Dz​=∣α2α3​23α​11β​∣=0

$$\Rightarrow {\alpha }^{2}-\alpha \beta -3=0$$⇒α2−αβ−3=0

Now when $\beta =2,\alpha =-1$β=2,α=−1 the system of equations will have infinite solutions,

When $\alpha =3,\beta =2$α=3,β=2 the system has infinite solution,

And when $\alpha =3,\beta \neq 2$α=3,β=2 the system has no solution as {D}_{x}&{D}_{y}\neq 0 for $\beta \neq 2$β=2,

Hence, only option $(B)$(B) is not correct