Let $p,q,r$ be non-zero real numbers that are, respectively, the ${10}^{\mathrm{th} },{100}^{\mathrm{th} }$ and ${1000}^{\mathrm{th} }$ terms of a harmonic progression. Consider the system of linear equations
$x+y+z=1$

$10x+100y+1000z=0$

$qrx+pry+pqz=0$

	List-$\mathrm{I}$		List-$\mathrm{II}$
$\left(\mathrm{I}\right)$	If $\frac{q}{r}=10$, then the system of linear equations has	$\left(\mathrm{P}\right)$	$x=0,y=\frac{10}{9},z=-\frac{1}{9}$ as a solution
$\left(\mathrm{II}\right)$	If $\frac{p}{r}\ne 100$, then the system of linear equations has	$\left(\mathrm{Q}\right)$	$x=\frac{10}{9},y=-\frac{1}{9},z=0$ as a solution
$\left(\mathrm{III}\right)$	If $\frac{p}{q}\ne 10$, then the system of linear equations has	$\left(\mathrm{R}\right)$	infinitely many solutions
$\left(\mathrm{IV}\right)$	If $\frac{p}{q}=10$, then the system of linear equations has	$\left(\mathrm{S}\right)$	no solution
		$\left(\mathrm{T}\right)$	at least one solution

The correct option is:

Given,

$x+y+z=1 \cdots \left(1\right)$

$10x+100y+1000z=0 \cdots \left(2\right)$

$qrx+pry+pqz=0 \cdots \left(3\right)$

Now equation $\left(3\right)$ can be re-written as

$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=0 \left(\because p,q,r\ne 0\right)$

Now given $p, q \& r{\text{ are 10}}^{th}{\text{, 100}}^{th}{\text{ \& 1000}}^{th}\text{ term of an h.p}$,

So, let $p=\frac{1}{a+9d}, q=\frac{1}{a+99d} \& r=\frac{1}{a+999d}$

Now, equation $\left(3\right)$ will be

$\left(a+9d\right)x+\left(a+99d\right)y+\left(a+999d\right)z=0$

Now from equation $\left(1\right), \left(2\right) \& \left(3\right)$ we get,

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 10& 100& 1000\\ a+9d& a+99d& a+999d\end{array}\right|=0$

${\Delta }_x=\left|\begin{array}{ccc}1& 1& 1\\ 0& 100& 1000\\ 0& a+99d& a+999d\end{array}\right|=900\left(d-a\right)$

${\Delta }_y=\left|\begin{array}{ccc}1& 1& 1\\ 10& 0& 1000\\ a+9d& 0& a+999d\end{array}\right|=990\left(a-d\right)$

${\Delta }_z=\left|\begin{array}{ccc}1& 1& 1\\ 10& 100& 0\\ a+9d& a+99d& 0\end{array}\right|=90\left(d-a\right)$

Option I: If $\frac{q}{r}=10\Rightarrow a=d$

$\Delta ={\Delta }_x={\Delta }_y={\Delta }_z=0$

And eq. $\left(1\right)$ and eq. $\left(2\right)$ represents non-parallel planes eq. $\left(2\right)$ and eq. $\left(3\right)$ represents same plane

$\Rightarrow$ Infinitely many solutions

Now finding solution by taking $z=\lambda$ so from equation $\left(1\right) \& \left(2\right)$ we get,

$x+y=1-\lambda$ and $x+10y=-100\lambda$

$\Rightarrow x=\frac{10}{9}+10\lambda , y=\frac{-1}{9}-11\lambda$

$\Rightarrow$$\left(x,y,z\right)\in \left(\frac{10}{9}+10\lambda , \frac{-1}{9}-11\lambda , \lambda \right)$

So, $P$ is not valid for any value of $\lambda$ rest are valid.

So, option (i)$\to Q,R,T$

Option II: $\frac{p}{r}\ne 100\Rightarrow a\ne d$

$\Delta =0 \& {\Delta }_x,{\Delta }_y,{\Delta }_z\ne 0$

So, no solution

Option (ii)$\to S$

Option (iii): If $\frac{p}{q}\ne 10\Rightarrow a\ne d$ then ${\Delta }_z\ne 0$

So, no solution

Option (iii) $\to S$

Now option (iv): If $\frac{p}{q}=10\Rightarrow a=d$ then ${\Delta }_z=0\Rightarrow {\Delta }_x={\Delta }_y=0$

So, infinitely many solutions

Option (iv)$\to Q,R,T$ {similar to option (i)}