If the values of $x, y$ and $z$ which satisfy the equations $2x - 3y + 2z + 15 = 0$,$3x + y - z + 2 = 0$ and $x - 3y - 3z + 8 = 0$ simultaneously are $\alpha, \beta$ and $\gamma$ respectively,then:

In matrix notation, if the system of equations $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$ has an infinite number of solutions, then all these solutions lie on

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$,$\lambda x+(\lambda-1) y+(\lambda-4) z=7$,and $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions,then $\lambda^2+\lambda$ is equal to

Let $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]$,$B=\left[B_1, B_2, B_3\right]$,where $B_1, B_2, B_3$ are column matrices,and $AB_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$,$AB_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right]$,$AB_3=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$. If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$,then $\alpha^3+\beta^3$ is equal to

If the system of equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method, then:

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