Let p, q, r be non-zero real numbers that are, respectively, the 10...

Given,

$x + y + z = 1 ⋯ \left(1\right)$

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

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

Now equation \left(3\right)$can be re-written as</p><p>$\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 0\left(∵ p, q, r \ne 0\right)$</p><p>Now given$p, q \text{ and } r{\text{are 10}}^{th}{\text{, 100}}^{th}{\text{\text{ and } 1000}}^{th}\text{term of an h.p}$,</p><p>So, let$p = \dfrac{1}{a + 9d}, q = \dfrac{1}{a + 99d} \text{ and } r = \dfrac{1}{a + 999d}$</p><p>Now, equation \left(3\right)$ will be

\left(a + 9d\right)x + \left(a + 99d\right)y + \left(a + 999d\right)z = 0$</p><p>Now from equation \left(1\right), \left(2\right) \text{ and } \left(3\right)$ we get,

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

{\Delta}_{x} = \left|\begin{cases}1 & 1 & 1 \\ 0 & 100 & 1000 \\ 0 & a + 99d & a + 999d\end{cases}\right| = 900\left(d - a\right)

{\Delta}_{y} = \left|\begin{cases}1 & 1 & 1 \\ 10 & 0 & 1000 \\ a + 9d & 0 & a + 999d\end{cases}\right| = 990\left(a - d\right)

{\Delta}_{z} = \left|\begin{cases}1 & 1 & 1 \\ 10 & 100 & 0 \\ a + 9d & a + 99d & 0\end{cases}\right| = 90\left(d - a\right)

Option I: If $\dfrac{q}{r} = 10 ⇒ a = d$

$\Delta = {\Delta}_{x} = {\Delta}_{y} = {\Delta}_{z} = 0$

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

$⇒$ Infinitely many solutions

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

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

$⇒ x = \dfrac{10}{9} + 10\lambda, y = \dfrac{- 1}{9} - 11\lambda$

$⇒$$\left(x, y, z\right) \in \left(\dfrac{10}{9} + 10\lambda, \dfrac{- 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: $\dfrac{p}{r} \ne 100 ⇒ a \ne d$

$\Delta = 0 \text{ and } {\Delta}_{x}, {\Delta}_{y}, {\Delta}_{z} \ne 0$

So, no solution

Option (ii)$\to S$

Option (iii): If $\dfrac{p}{q} \ne 10 ⇒ a \ne d$ then ${\Delta}_{z} \ne 0$

So, no solution

Option (iii) $\to S$

Now option (iv): If $\dfrac{p}{q} = 10 ⇒ a = d$ then ${\Delta}_{z} = 0 ⇒ {\Delta}_{x} = {\Delta}_{y} = 0$

So, infinitely many solutions

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