If the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$,and $x + y + 3z = 4$ has infinitely many solutions,then $p=$

If $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right]$,$X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$,then find the value of $2x+y-z$.

Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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 system of linear equations $x + ky + 3z = 0$,$3x + ky - 2z = 0$,and $2x + 4y - 3z = 0$ has a non-zero solution $(x, y, z)$,then $\frac{xz}{y^2} = \dots$

If the system of equations $kx + 2y - z = 2, (k - 1)x + ky + z = 1, x + (k - 1)y + kz = 3$ has only one solution,then the number of possible real value$(s)$ of $k$ is -