1. Use the property of a harmonic progression

If $p,q,r$ are the $10^{\text{th}},100^{\text{th}},1000^{\text{th}}$ terms of an H.P., then their reciprocals are corresponding terms of an A.P.

So,

$$\frac1p,\; \frac1q,\; \frac1r$$

are the $10^{\text{th}},100^{\text{th}},1000^{\text{th}}$ terms of some A.P.

Let the A.P. have first term $a$ and common difference $d$. Then

$$\frac1p=a+9d,\qquad \frac1q=a+99d,\qquad \frac1r=a+999d.$$

Hence,

$$\frac1q-\frac1p=90d,\qquad \frac1r-\frac1q=900d.$$

Therefore,

$$\frac1r-\frac1q=10\left(\frac1q-\frac1p\right).$$

This relation may be useful, but the main work is with the system.

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2. Rewrite the third equation

Given system:

$$x+y+z=1 \tag{1}$$ $$10x+100y+1000z=0 \tag{2}$$ $$qrx+pry+pqz=0. \tag{3}$$

Since $p,q,r\neq 0$, divide equation (3) by $pqr$:

$$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=0. \tag{3'}$$

So the coefficient matrix is

$$\begin{pmatrix} 1&1&1\\ 10&100&1000\\ \frac1p&\frac1q&\frac1r \end{pmatrix}.$$

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3. Observe when the third row becomes dependent on the first two

Suppose

$$\frac1p,\frac1q,\frac1r$$

lie on a line in the index variable. Since these are terms of an A.P., there exist constants $\alpha,\beta$ such that

$$\frac1p=\alpha+10\beta,\qquad \frac1q=\alpha+100\beta,\qquad \frac1r=\alpha+1000\beta.$$

Thus the third row is a linear combination of the first two rows. Hence determinant is always $0$.

So the system can never have a unique solution; it will have either no solution or infinitely many solutions.

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4. Find the solutions of the first two equations

From

$$x+y+z=1$$

and

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

we solve for special cases.

Case A: $z=0$

Then

$$x+y=1,$$ $$10x+100y=0 \Rightarrow x+10y=0.$$

Subtracting,

$$(x+10y)-(x+y)=0-1 \Rightarrow 9y=-1 \Rightarrow y=-\frac19,$$

so

$$x=\frac{10}{9}.$$

Hence

$$\left(\frac{10}{9},-\frac19,0\right)$$

is a solution of the first two equations. This is exactly (Q).

Case B: $x=0$

Then

$$y+z=1,$$ $$100y+1000z=0 \Rightarrow y+10z=0.$$

Subtracting,

$$(y+10z)-(y+z)=0-1 \Rightarrow 9z=-1 \Rightarrow z=-\frac19,$$

so

$$y=\frac{10}{9}.$$

Hence

$$\left(0,\frac{10}{9},-\frac19\right)$$

is a solution of the first two equations. This is exactly (P).

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5. Check each statement of List-I

We use equation

$$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=0.$$

(I) If $\dfrac{q}{r}=10$

Then

$$\frac1q=\frac{10}{r}=10\cdot\frac1r.$$

Equivalently,

$$\frac1r=\frac1{10q}.$$

But easiest is to test option (Q):

$$(x,y,z)=\left(\frac{10}{9},-\frac19,0\right).$$

Then third equation becomes

$$\frac{1}{p}\cdot\frac{10}{9}+\frac{1}{q}\cdot\left(-\frac19\right)=0$$ $$\Rightarrow \frac{10}{p}=\frac{1}{q}$$ $$\Rightarrow \frac{p}{q}=10.$$

This is not given.

Now test (P):

$$(x,y,z)=\left(0,\frac{10}{9},-\frac19\right).$$

Then third equation becomes

$$\frac{10}{9q}-\frac{1}{9r}=0$$ $$\Rightarrow \frac{10}{q}=\frac{1}{r}$$ $$\Rightarrow q=10r$$ $$\Rightarrow \frac{q}{r}=10.$$

So under condition (I), (P) is indeed a solution, hence certainly the system has at least one solution. Thus

$$(I)\to (T).$$

So any option with $(I)\to(Q)$ is false.

Therefore B and C are immediately suspect. But let us continue fully.

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(II) If $\dfrac{p}{r}\ne 100$

Consider consistency of third equation with first two.

Because the third row is dependent on first two rows, consistency fails exactly when the same linear combination on RHS does not match.

Let

$$\frac1p=\alpha+10\beta,\quad \frac1q=\alpha+100\beta,\quad \frac1r=\alpha+1000\beta.$$

Subtract first from second:

$$\frac1q-\frac1p=90\beta.$$

Subtract second from third:

$$\frac1r-\frac1q=900\beta.$$

Hence true.

Now if third row equals $\alpha$(row 1)$+\beta$(row 2), then for consistency RHS must satisfy

$$0=\alpha\cdot 1+\beta\cdot 0=\alpha.$$

So we need $\alpha=0$ for consistency.

If $\alpha=0$, then

$$\frac1p: \frac1q: \frac1r = 10:100:1000 = 1:10:100,$$

so

$$p:q:r = 100:10:1.$$

Therefore

$$\frac{p}{r}=100,$$ $$\frac{p}{q}=10,$$ $$\frac{q}{r}=10.$$

Thus the system is consistent iff $\dfrac{p}{r}=100$. So if

$$\frac{p}{r}\ne 100,$$

then the system has no solution. Hence

$$(II)\to (S).$$

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(III) If $\dfrac{p}{q}\ne 10$

From above, consistency requires

$$\frac{p}{q}=10$$

(as part of the same ratio condition). So if

$$\frac{p}{q}\ne 10,$$

then system is inconsistent. Hence

$$(III)\to (S).$$

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(IV) If $\dfrac{p}{q}=10$

Since $p,q,r$ are from the same H.P., let

$$\frac1p=a+9d,\qquad \frac1q=a+99d.$$

Condition $\dfrac{p}{q}=10$ means

$$\frac{1/q}{1/p}=10 \Rightarrow \frac1q=10\cdot \frac1p.$$

So

$$a+99d=10(a+9d)$$ $$a+99d=10a+90d$$ $$9d=9a$$ $$a=d.$$

Then

$$\frac1r=a+999d=1000d.$$

Also

$$\frac1p=a+9d=10d,$$ $$\frac1q=a+99d=100d.$$

Thus

$$\frac1p:\frac1q:\frac1r = 10:100:1000,$$

so the third row is exactly a scalar multiple of the second row:

$$\left(\frac1p,\frac1q,\frac1r\right)=d(10,100,1000).$$

Since equation (2) has RHS $0$, equation (3') is then dependent on equation (2), and the system reduces to two independent equations in three variables. Hence there are infinitely many solutions. Thus

$$(IV)\to (R).$$

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6. Final matching

We obtained:

$$(I)\to (T),\qquad (II)\to (S),\qquad (III)\to (S),\qquad (IV)\to (R).$$

This corresponds to Option D? Check carefully:

• A: $(I)\to T$, $(II)\to R$, $(III)\to S$, $(IV)\to T$ ❌
• B: $(I)\to Q$, $(II)\to S$, $(III)\to S$, $(IV)\to R$ ❌ because $(I)\to Q$ is false
• C: $(I)\to Q$, $(II)\to R$, $(III)\to P$, $(IV)\to R$ ❌
• D: $(I)\to T$, $(II)\to S$, $(III)\to P$, $(IV)\to T$ ❌ because $(III)$ should be $S$ and $(IV)$ should be $R$

So none of the given options matches all correct relations.

However, among the listed assertions individually:

• $(I)\to T$ is true,
• $(II)\to S$ is true,
• $(III)\to S$ is true,
• $(IV)\to R$ is true.

Thus the intended correct mapping is

$$(I)\to(T),\ (II)\to(S),\ (III)\to(S),\ (IV)\to(R).$$

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7. Compare with stored answer

Stored answer is B, but option B says $(I)\to(Q)$, which is incorrect. Under $\dfrac{q}{r}=10$, the specific solution is (P), and certainly (T) is true, not (Q).

Therefore, I disagree with the stored answer.