If the system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = 0$ has a unique solution,then $\lambda$ is not equal to:

If the system of equations $3x + y + 4z = 3$,$2x + ay - z = -3$,$x + 2y + z = 4$ has no solution,then the value of $a$ is equal to:

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:

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 the system of equations $x+2y+3z=3$,$4x+3y-4z=4$,and $8x+4y-\lambda z=9+\mu$ has infinitely many solutions,then the ordered pair $(\lambda, \mu)$ 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: