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(D) The given system of equations is:$(i) \ 2x + y - z = 3$$(ii) \ x - y - z = \alpha$$(iii) \ 3x + 3y + \beta z = 3$For a system of linear equations

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$5$

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(D) The given system of equations is:$(i) \ 2x + y - z = 3$$(ii) \ x - y - z = \alpha$$(iii) \ 3x + 3y + \beta z = 3$For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must be consistent.Let the coefficient matrix be $A = \begin{bmatrix} 2 & 1 & -1 \ 1 & -1 & -1 \ 3 & 3 & \beta \end{bmatrix}$.Setting $|A| = 0$:$2(-\beta + 3) - 1(\beta + 3) - 1(3 + 3) = 0$$-2\beta + 6 - \beta - 3 - 6 = 0$$-3\beta - 3 = 0 \Rightarrow \beta = -1$.Substituting $\beta = -1$ into the system:$2x + y - z = 3$$x - y - z = \alpha$$3x + 3y - z = 3$Subtracting $(ii)$ from $(i)$ gives $x + 2y = 3 - \alpha$.From $(iii)$,$3(x + y) - z = 3$. Using $(i) + (ii)$,$3x = 3 + \alpha$,so $x = 1 + \alpha/3$.For consistency,the equations must represent the same plane or line. Solving the system leads to $\alpha = 3$.Thus,$\alpha + \beta - \alpha \beta = 3 + (-1) - (3)(-1) = 3 - 1 + 3 = 5$.

If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .

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If the system of linear equations$2x + y - z = 3$$x - y - z = \alpha$$3x + 3y + \beta z = 3$has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .