The set of equations $x - y + 3z = 2$,$2x - y + z = 4$,and $x - 2y + \alpha z = 3$ has:

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 $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4 \end{bmatrix}$,$B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ and $X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ such that $AX = B$,then the value of $x_1 + x_2 + x_3$ is:

The number of real values of $\alpha$ for which the system of equations
$x+3y+5z=\alpha x$
$5x+y+3z=\alpha y$
$3x+5y+z=\alpha z$
has infinite number of solutions is

If $\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$,then $X =$

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