If the system of linear equations $2x + 3y - z = -2$; $x + y + z = 4$; $x - y + |\lambda|z = 4\lambda - 4$ (where $\lambda \in R$) has no solution,then:

The system of equations $x+y+z=5$, $x+2y+3z=9$ and $x+3y+\lambda z=\mu$ has a unique solution if

Examine the consistency of the system of equations: $x+y+z=1$,$2x+3y+2z=2$,and $ax+ay+2az=4$.

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 equations $2x - y + z = 4$,$5x + \lambda y + 3z = 12$,and $100x - 47y + \mu z = 212$ has infinitely many solutions,then $\mu - 2\lambda$ is equal to

If $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix}$,then $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is equal to