If $x=\alpha, y=\beta, z=\gamma$ is the unique solution of the system of linear equations $2x-3y+5z=12$, $5x+2y+3z=11$, and $x+2y-3z=-3$, then $2\alpha+5\beta+3\gamma=$

If the system of equations $x+y+z=6$,$2x+5y+\alpha z=\beta$,and $x+2y+3z=14$ has infinitely many solutions,then $\alpha+\beta$ is equal to.

The values of $x, y, z$ in order for the system of equations $3x + y + 2z = 3,$ $2x - 3y - z = -3,$ and $x + 2y + z = 4$ are:

Let $p, q, r$ be nonzero real numbers that are,respectively,the $10^{\text{th}}$,$100^{\text{th}}$,and $1000^{\text{th}}$ terms of a harmonic progression. Consider the system of linear equations:
$x+y+z=1$
$10x+100y+1000z=0$
$qrx + pry + pqz = 0$

$List-I$	$List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has	$(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has	$(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has	$(S)$ no solution
	$(T)$ at least one solution

The correct option is:

If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$,find $A^{-1}$. Using $A^{-1}$,solve the system of equations: $2x - 3y + 5z = 11$,$3x + 2y - 4z = -5$,and $x + y - 2z = -3$.

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