If the system of equations $x+y+z=6$,$2x+5y+\alpha z=\beta$,and $x+2y+3z=14$ has infinitely many solutions,then $\alpha+\beta$ is equal to.

For the system of linear equations:
$2x - y + 3z = 5$
$3x + 2y - z = 7$
$4x + 5y + \alpha z = \beta$
Which of the following is $NOT$ correct?

If $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$,$B=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ and $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,then $x+y+z=$

The solution of the equation $\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ is $(x, y, z) = $

The values of $p$ and $q$ so that the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$ and $x + y + 3z = 4$ may have no solution are

The value of $\lambda$ such that the system of equations $2x-y-2z=2$,$x-2y+z=-4$,and $x+y+\lambda z=4$ has no solution,is: