For the system of linear equations x + y + z = 6 + + 7z = 3 x + 2y...

Given system of linear equations:

$x + y + z = 6$

$\alphax + \betay + 7z = 3$

$x + 2y + 3z = 14$

On taking determinant, we get

\left|A\right| = \left|$\begin{cases}1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3\end{cases}$\right|$</p><p>$= ( 3\beta - 14 ) - ( 3\alpha - 7 ) + ( 2\alpha - \beta )$</p><p>$= 2\beta - \alpha - 7$</p><p>Here, if$\alpha = \beta = 7 then \left|A\right| = 0$and the system has no solution. So, option A is correct.</p><p>If$\alpha = \beta, \alpha \ne 7$then the system has a unique solution. So, option B is correct.</p><p>If$( \alpha, \beta ) \ne ( 7, 7 )$then \left|A\right| \ne 0$. So, system has not infinitely many solution. Therefore, option $D$ is false.