Solve the system of linear equations using the matrix method: $x-y+2z=7$,$3x+4y-5z=-5$,$2x-y+3z=12$.

If the system of equations $x + y + z = 5$, $x + 2y + 3z = 9$, $x + 3y + \lambda z = \mu$ has infinitely many solutions, then the value of $\lambda + \mu$ is:

Let $M = (a_{ij})$,$i, j \in \{1, 2, 3\}$,be a $3 \times 3$ matrix such that $a_{ij} = 1$ if $j+1$ is divisible by $i$,otherwise $a_{ij} = 0$. Then which of the following statements is (are) true?
$(A)$ $M$ is invertible
$(B)$ There exists a nonzero column matrix $\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$ such that $M \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} -a_1 \\ -a_2 \\ -a_3 \end{bmatrix}$
$(C)$ The set $\{X \in \mathbb{R}^3 : MX = 0, X \neq 0\}$ is non-empty,where $0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
$(D)$ The matrix $(M - 2I)$ is invertible,where $I$ is the $3 \times 3$ identity matrix

The values of $a$ and $b$,for which the system of equations $2x + 3y + 6z = 8$,$x + 2y + az = 5$,and $3x + 5y + 9z = b$ has no solution,are:

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:

Let $a, b, c, d, e$ be five numbers satisfying the system of equations:
$2a + b + c + d + e = 6$
$a + 2b + c + d + e = 12$
$a + b + 2c + d + e = 24$
$a + b + c + 2d + e = 48$
$a + b + c + d + 2e = 96$
Then $|c|$ is equal to: