If the system of equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method, then:

For how many values of $k$ does the system of linear equations $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$ have no solutions?

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

If the values of $x, y$ and $z$ which satisfy the equations $2x - 3y + 2z + 15 = 0$,$3x + y - z + 2 = 0$ and $x - 3y - 3z + 8 = 0$ simultaneously are $\alpha, \beta$ and $\gamma$ respectively,then:

If the system of equations $\begin{bmatrix} \alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \alpha-1 \\ \alpha-1 \\ \alpha-1 \end{bmatrix}$ is inconsistent, then $\alpha=$

The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is