If the system of linear equations $x + 2ay + az = 0$,$x + 3by + bz = 0$,and $x + 4cy + cz = 0$ has a non-zero solution,then $a, b, c$:

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 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

For what value of $\lambda$,the system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = 12$ is inconsistent? $\lambda = $ ........

The number of non-trivial solutions of the system $x-y+z=0, x+2y-z=0, 2x+y+3z=0$ is

The system of equations $3x + 2y + z = 6$, $3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to