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(B) For a system of linear equations to have infinite solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\

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$5$

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(B) For a system of linear equations to have infinite solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\Delta_1, \Delta_2, \Delta_3$ must also be $0$.First,calculate $\Delta$:$\Delta = \begin{vmatrix} 1 & 1 & -1 \ 1 & 2 & \alpha \ 2 & -1 & 1 \end{vmatrix} = 1(2 + \alpha) - 1(1 - 2\alpha) - 1(-1 - 4) = 2 + \alpha - 1 + 2\alpha + 5 = 3\alpha + 6$.Setting $\Delta = 0$,we get $3\alpha + 6 = 0$,which implies $\alpha = -2$.Now,substitute $\alpha = -2$ into the system:$x + y - z = 2$$x + 2y - 2z = 1$$2x - y + z = \beta$For infinite solutions,the augmented matrix must have a rank less than $3$. We can use Cramer's rule or row reduction. Using $\Delta_2 = 0$:$\Delta_2 = \begin{vmatrix} 1 & 2 & -1 \ 1 & 1 & -2 \ 2 & \beta & 1 \end{vmatrix} = 1(1 + 2\beta) - 2(1 + 4) - 1(\beta - 2) = 1 + 2\beta - 10 - \beta + 2 = \beta - 7$.Setting $\Delta_2 = 0$,we get $\beta = 7$.Thus,$\alpha + \beta = -2 + 7 = 5$.

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

For real numbers $\alpha$ and $\beta$,consider the following system of linear equations:
$x+y-z=2, x+2y+\alpha z=1, 2x-y+z=\beta$. If the system has infinite solutions,then $\alpha+\beta$ is equal to $.....$

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For real numbers $\alpha$ and $\beta$,consider the following system of linear equations:$x+y-z=2, x+2y+\alpha z=1, 2x-y+z=\beta$. If the system has infinite solutions,then $\alpha+\beta$ is equal to $.....$