If the system of simultaneous linear equations $x+y-z=6$,$4x+y+z=2$,and $x+ky+z=-8$ has a unique solution $x=2$,$y=\beta$,$z=\gamma$,then the value of $k$ satisfies which of the following quadratic equations?

Let $M$ be a $3 \times 3$ matrix such that $M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$,$M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$ and $M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$. If $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$,then $x + y + z$ equals :

The system of linear equations $x + y + z = 2, 2x + 3y + 2z = 5$,and $2x + 3y + (a^2 - 1)z = a + 1$ is:

$A, C$ are $3 \times 3$ matrices. $B, D$ are $3 \times 1$ matrices. If $AX=B$ has a unique solution and $CX=D$ has an infinite number of solutions,then:

Use the product $\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{lll}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations:
$x-y+2z=1$
$2y-3z=1$
$3x-2y+4z=2$

If the system of linear equations: $x + y + z = 6, x + 2y + 5z = 10, 2x + 3y + \lambda z = \mu$ has infinitely many solutions,then the value of $\lambda + \mu$ equals: