Let $x^{pq^2}=y^{qr}=z^{p^2r}$ and $r=pq+1,$ where $p,q,r$ are positive integers.

Also, $3,\, 3\log_y x,\, 3\log_z y,\, 7\log_x z$ are in A.P. with common difference $\frac12$.

We solve step by step.

1. Use the A.P. condition

Since the four numbers are in A.P. with first term $3$ and common difference $\frac12$,

$3\log_y x = 3+\frac12=\frac72,$ $3\log_z y = 3+2\cdot\frac12=4,$ $7\log_x z = 3+3\cdot\frac12=\frac92.$

Hence, $\log_y x=\frac76,$ $\log_z y=\frac43,$ $\log_x z=\frac{9}{14}.$

2. Convert these logarithms into relations among $x,y,z$

From $\log_y x=\frac76,$ we get $x=y^{7/6}.$

From $\log_z y=\frac43,$ we get $y=z^{4/3}.$

From $\log_x z=\frac{9}{14},$ we get $z=x^{9/14}.$

These are consistent, since $x=y^{7/6}=(z^{4/3})^{7/6}=z^{14/9},$ so indeed $z=x^{9/14}.$

3. Use the given equal powers

Let $x^{pq^2}=y^{qr}=z^{p^2r}=k.$

Take logs in a convenient way using the known logarithmic relations.

From $x^{pq^2}=y^{qr}$

Since $x=y^{7/6}$, $(y^{7/6})^{pq^2}=y^{qr}.$ Thus, $\frac76 pq^2=qr.$ Since $q>0$, $\frac76 pq=r.$ So, $r=\frac76 pq. \qquad (1)$

From $y^{qr}=z^{p^2r}$

Since $y=z^{4/3}$, $(z^{4/3})^{qr}=z^{p^2r}.$ Thus, $\frac43 qr=p^2r.$ Since $r>0$, $\frac43 q=p^2. \qquad (2)$

4. Use $r=pq+1$

From (1), $r=\frac76 pq.$ But also given, $r=pq+1.$ Therefore, $\frac76 pq=pq+1.$ So, $\left(\frac76-1\right)pq=1$ $\frac16 pq=1$ $pq=6.$

Then $r=pq+1=7.$

5. Find $p$ and $q$

From (2), $p^2=\frac43 q.$ Since $pq=6$, we test factor pairs of $6$:

• $(p,q)=(1,6)$ gives $p^2=1$, $\frac43 q=8$ ❌
• $(p,q)=(2,3)$ gives $p^2=4$, $\frac43 q=4$ ✅
• $(p,q)=(3,2)$ gives $p^2=9$, $\frac43 q=\frac83$ ❌
• $(p,q)=(6,1)$ gives $p^2=36$, $\frac43 q=\frac43$ ❌

Hence, $p=2,\quad q=3,\quad r=7.$

6. Compute $r-p-q$

$r-p-q=7-2-3=2.$

7. Compare with the stored answer

Derived answer is: $\boxed{2}$ This matches option D.

So the stored correct answer is consistent.