For the equations $x+2y+3z=1$,$2x+y+3z=2$,and $5x+5y+9z=4$,which of the following is true?

For what values of $a$ will the system of equations $x+y+z=1$, $2x+3y+2z=2$, and $ax+ay+2az=4$ have a unique solution?

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

Let $A$ be a $3 \times 3$ real matrix such that $A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$,$A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$. If $X = (x_1, x_2, x_3)^T$ and $I$ is an identity matrix of order $3$,then the system $(A - 2I)X = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$ has:

If the system of linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

Let $X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$,$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix}$,and $B = \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}$. If $AX = B$,then the value of $2a - 3b + 4c$ is: