Let $$p$$, $$q$$, $$r$$ be nonzero real numbers that are, respectively, the $$10^{th}$$, $$100^{th}$$ and $$1000^{th}$$ terms of a harmonic progression. Consider the system of linear equations

$$x + y + z = 1$$

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

$$qr \, x + pr \, y + pq \, z = 0.$$

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:

Let the reciprocals of the terms of the harmonic progression be in an arithmetic progression with first term $$a$$ and common difference $$d$$.
Then the $$n^{\text{th}}$$ term of the HP is $$\displaystyle T_n=\frac{1}{a+(n-1)d}$$.

Hence
$$p=\frac{1}{u},\;q=\frac{1}{v},\;r=\frac{1}{w},$$
where $$u=a+9d,\;v=a+99d,\;w=a+999d$$  $$-(1)$$

The given system is
$$x+y+z=1$$ $$-(2)$$
$$10x+100y+1000z=0\Longrightarrow x+10y+100z=0$$ $$-(3)$$
$$qr\,x+pr\,y+pq\,z=0$$ $$-(4)$$

Step 1 : Solution of the first two equations
Subtract $$(2)$$ from the simplified $$(3)$$:

$$9y+99z=-1\;\Longrightarrow\;y=-\frac19-11z\;$$ $$-(5)$$
Putting this in $$(2)$$:

$$x=1-\bigl(-\tfrac19-11z\bigr)-z=\frac{10}{9}+10z$$ $$-(6)$$

Thus every common solution of $$(2),(3)$$ can be written as
$$x=\frac{10}{9}+10t,\;y=-\frac19-11t,\;z=t\quad(t\in\mathbb{R})$$ $$-(7)$$

Step 2 : Using the third equation
Substitute $$(7)$$ in $$(4)$$:

$$qr\Bigl(\frac{10}{9}+10t\Bigr)+pr\Bigl(-\frac19-11t\Bigr)+pq\,t=0$$

Multiply by $$9$$ and collect the constant and the $$t$$ terms:

$$r(10q-p)+t\,(10qr-11pr+pq)=0$$ $$-(8)$$

Step 3 : Express the coefficients in terms of $$u,v,w$$
From $$(1)$$ we get

$$qr=\frac1{vw},\;pr=\frac1{uw},\;pq=\frac1{uv}$$

Multiplying the left side of $$(8)$$ by $$9uvw$$ gives

$$10u-v+ t\,(10u-11v+w)=0$$ $$-(9)$$

Now compute the combination in the bracket using $$u=a+9d,\;v=a+99d,\;w=a+999d$$:

$$10u-11v+w=10(a+9d)-11(a+99d)+(a+999d)=0$$

Therefore
$$10u-11v+w\equiv0\quad\text{for all }a,d$$ $$-(10)$$

Equation $$(9)$$ reduces to

$$\bigl(10u-v\bigr)=0\quad\text{or}\quad10u-v\neq0$$

Introduce the useful ratios
$$\frac{p}{q}=\frac{v}{u},\qquad\frac{q}{r}=\frac{w}{v},\qquad\frac{p}{r}=\frac{w}{u}$$ $$-(11)$$

From $$(10)$$ we already know that the coefficient of $$t$$ is zero; hence:

• If $$10u-v=0$$
$$v=10u\;\Longrightarrow\;\frac{p}{q}=10,\;\frac{q}{r}=10,\;\frac{p}{r}=100$$ (put $$a=d$$ in $$(1)$$).
Then $$(8)$$ is satisfied for every $$t$$, giving infinitely many solutions.

• If $$10u-v\neq0$$
$$\displaystyle\frac{p}{q}\neq10,\;\frac{p}{r}\neq100,\;\frac{q}{r}\neq10$$.
Because the coefficient of $$t$$ is already zero, the constant term cannot also be zero. Hence $$(8)$$ is impossible and the system has no solution.

Step 4 : Special vectors
If $$10u-v=0$$ (that is, $$\tfrac{q}{r}=10$$), choose $$t=0$$ in $$(7)$$ to obtain the particular solution

$$x=\frac{10}{9},\;y=-\frac{1}{9},\;z=0$$ $$-(12)$$

Similarly, picking $$t=-1$$ gives $$x=0,\;y=\tfrac{10}{9},\;z=-\tfrac{1}{9}$$, etc. Thus both vectors listed in List-II actually satisfy the system whenever $$\tfrac{q}{r}=10$$.

Step 5 : Matching each statement

(I) $$\displaystyle\frac{q}{r}=10\;\Longrightarrow\;10u-v=0$$, so the system is consistent and the triple $$(12)$$ is indeed a solution. Hence (I) → (Q).

(II) $$\displaystyle\frac{p}{r}\neq100\;\Longrightarrow\;10u-v\neq0$$, so the system is inconsistent. Hence (II) → (S).

(III) $$\displaystyle\frac{p}{q}\neq10\;\Longrightarrow\;10u-v\neq0$$, giving no solution. Hence (III) → (S).

(IV) $$\displaystyle\frac{p}{q}=10\;\Longrightarrow\;10u-v=0$$, which yields infinitely many solutions. Hence (IV) → (R).

Therefore the correct set of connections is
$$\boxed{(I)\to(Q),\;(II)\to(S),\;(III)\to(S),\;(IV)\to(R)}$$

Option B matches this correspondence.