Consider the following system of questions + 2y + z = 1 2 + 3y + z...

Given,

System of questions

$\alphax + 2y + z = 1$

$2\alphax + 3y + z = 1$

$3x + \alphay + 2z = \beta$

Now finding, D = \left|\begin{cases}\alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2\end{cases}\right| = 0

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

$⇒ \left(\alpha - 3\right) ( \alpha + 1 ) = 0$

$⇒ \alpha = - 1, 3$

{D}_{x} = \left|\begin{cases}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{cases}\right| = 0 ⇒ \beta = 2

$⇒ 2\left(\beta - 2\right) - 1 ( 3\beta - \alpha ) + ( 6 - \alpha ) = 0$

$⇒ \beta = 2$

{D}_{y} = \left|\begin{cases}\alpha & 1 & 1 \\ 2\alpha & 1 & 1 \\ 3 & 2 & \beta\end{cases}\right| = 0

$⇒ \alpha ( \beta - 2 ) - 1 ( 2\alpha\beta - 3 ) + 1 ( 4\alpha - 3 ) = 0$

$⇒ \alpha ( 2 - \beta ) = 0$

$⇒ \alpha = 0, \beta = 2$

{D}_{z} = \left|\begin{cases}\alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & \beta\end{cases}\right| = 0

$⇒ {\alpha}^{2} - \alpha\beta - 3 = 0$

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

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

And when $\alpha = 3, \beta \ne 2$ the system has no solution as ${D}_{x} \text{ and } {D}_{y} \ne 0$ for $\beta \ne 2$,

Hence, only option $\left(B\right)$ is not correct