For the system of linear equations ax + y + z = 1 , x + ay + z = 1,...

For infinite solution $∆ = {∆}_{x} = {∆}_{y} = {∆}_{z} = 0$

∆ = \left|\begin{cases}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{cases}\right| = 0

$\alpha\left({\alpha}^{2} - 1\right) - 1\left(\alpha - 1\right) + 1\left(1 - \alpha\right) = 0$

$⇒ {\alpha}^{3} - 3\alpha + 2 = 0$

$⇒ {\alpha}^{2}\left(\alpha - 1\right) + \alpha\left(\alpha - 1\right) - 2\left(\alpha - 1\right) = 0$

$⇒ \left(\alpha - 1\right)\left({\alpha}^{2} + \alpha - 2\right) = 0$

$⇒ \alpha = 1, \alpha = - 2, 1$

If $\beta = 1$, then all planes are overlapping

$∴$ option (d) is correct.

Put $\alpha = 2, \beta = 1$ the system equations as,

$2x + y + z = 1$

$x + 2y + z = 1$

$x + y + 2z = 1$

Adding all three above equations we get,

$x + y + z = \dfrac{3}{4}$

$∴$ option (c) is correct.

Now, if $\alpha = - 2, \beta = 1$ then,

$∆ = 0, {∆}_{x} \ne 0$ (No solution)

$∴$ option (b) is correct.

Now, if $\alpha = 2$

$∆ \ne 0$ (unique solution exists)

$∴$ option (a) is incorrect