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(B) The given system of equations is:$x+y-z=6$  $(1)$$3x+2y-z=5$  $(2)$$2x-y-2z=-3$  $(3)$Using the augmented matrix method:$[A:B] = \begin{bmatrix} 1

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(B) The given system of equations is:$x+y-z=6$  $(1)$$3x+2y-z=5$  $(2)$$2x-y-2z=-3$  $(3)$Using the augmented matrix method:$[A:B] = \begin{bmatrix} 1 & 1 & -1 & | & 6 \ 3 & 2 & -1 & | & 5 \ 2 & -1 & -2 & | & -3 \end{bmatrix}$Applying row operations:$R_2 ightarrow R_2 - 3R_1$ and $R_3 ightarrow R_3 - 2R_1$:$[A:B] = \begin{bmatrix} 1 & 1 & -1 & | & 6 \ 0 & -1 & 2 & | & -13 \ 0 & -3 & 0 & | & -15 \end{bmatrix}$$R_3 ightarrow R_3 - 3R_2$:$[A:B] = \begin{bmatrix} 1 & 1 & -1 & | & 6 \ 0 & -1 & 2 & | & -13 \ 0 & 0 & -6 & | & 24 \end{bmatrix}$From the third row: $-6z = 24 \Rightarrow z = -4 = \gamma$.From the second row: $-y + 2z = -13 \Rightarrow -y + 2(-4) = -13 \Rightarrow -y - 8 = -13 \Rightarrow y = 5 = \beta$.From the first row: $x + y - z = 6 \Rightarrow x + 5 - (-4) = 6 \Rightarrow x + 9 = 6 \Rightarrow x = -3 = \alpha$.Therefore,$\alpha + \beta = -3 + 5 = 2$.

If the solution of the system of simultaneous linear equations $x+y-z=6$,$3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha+\beta=$

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