If $x=\alpha, y=\beta, z=\gamma$ is the solution of the system of equations $2x+3y+z=-1$, $3x+y+z=4$, and $x-3y-2z=1$, then the value of $\beta$ is:

The number of $3 \times 3$ matrices $A$ whose entries are either $0$ or $1$ and for which the system $A\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ has exactly two distinct solutions,is

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?

For the system $x-y+z=4, 2x+y-3z=0, x+y+z=2$,the values of $x, y, z$ respectively are given by

Let $A = \{X = (x, y, z)^{T} : PX = 0 \text{ and } x^{2} + y^{2} + z^{2} = 1\}$ where $P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}$,then the set $A$:

Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{*}, y^{*}, z^{*})$. If $(\alpha, x^{*}), (y^{*}, \alpha)$ and $(x^{*}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is