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:

Consider the following system of equations: $x+2y-3z=a$,$2x+6y-11z=b$,and $x-2y+7z=c$,where $a, b$,and $c$ are real constants. Then the system of equations:

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 $x + 2y - 3z = 1$,$(k + 3)z = 3$,and $(2k + 1)x + z = 0$ is inconsistent,then the value of $k$ is

Consider the system of equations: $x + ay = 0$,$y + az = 0$,and $z + ax = 0$. The set of all real values of $a$ for which the system has a unique solution is:

The number of real values of $\lambda$ for which the system of linear equations $2x + 4y - \lambda z = 0$,$4x + \lambda y + 2z = 0$,and $\lambda x + 2y + 2z = 0$ has infinitely many solutions is: