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 -

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:

The system of linear equations $x + y + z = 2, 2x + 3y + 2z = 5$,and $2x + 3y + (a^2 - 1)z = a + 1$ is:

Under which of the following condition$(s)$ does the system of equations $\begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & a-4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \\ a \end{bmatrix}$ possess a unique solution?

If $AX=B$,where $A=\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix}$,$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B=\begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$,then $x^{2}+y^{2}+z^{2}=$

The sum of three numbers is $6$. If we multiply the third number by $3$ and add the second number to it,we get $11$. By adding the first and third numbers,we get double the second number. Represent this algebraically and find the numbers using the matrix method.