The system of equations $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = -18$ has

Let $n$ be the number obtained on rolling a fair die. If the probability that the system of equations
$x-ny+z=6$
$x+(n-2)y+(n+1)z=8$
$(n-1)y+z=1$
has a unique solution is $\frac{k}{6}$,then the sum of $k$ and all possible values of $n$ is:

If the system of equations $2x - y + z = 4$,$5x + \lambda y + 3z = 12$,and $100x - 47y + \mu z = 212$ has infinitely many solutions,then $\mu - 2\lambda$ 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:

The following system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,$x + 2y - 3z = 0$ has a solution other than $x = y = z = 0$ for $\lambda$ equal to

For the system of linear equations $x+y+z=6$; $\alpha x+\beta y+7z=3$; $x+2y+3z=14$,which of the following is $NOT$ true?