For three positive integers $$p, q, r$$, $$x^{pq^2} = y^{qr} = z^{p^2r}$$ and $$r = pq + 1$$ such that $$3, 3\log_y x, 3\log_z y, 7\log_x z$$ are in A.P. with common difference $$\frac{1}{2}$$. The $$r - p - q$$ is equal to

Given: $$x^{pq^2} = y^{qr} = z^{p^2r}$$ and $$r = pq + 1$$, with $$3, 3\log_y x, 3\log_z y, 7\log_x z$$ in AP with common difference $$d = 1/2$$.

Find the logarithmic relationships.

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

$$x = k^{1/(pq^2)}, \quad y = k^{1/(qr)}, \quad z = k^{1/(p^2r)}$$

Computing the log ratios:

$$\log_y x = \frac{\ln x}{\ln y} = \frac{1/(pq^2)}{1/(qr)} = \frac{qr}{pq^2} = \frac{r}{pq}$$ $$\log_z y = \frac{\ln y}{\ln z} = \frac{1/(qr)}{1/(p^2r)} = \frac{p^2r}{qr} = \frac{p^2}{q}$$ $$\log_x z = \frac{\ln z}{\ln x} = \frac{1/(p^2r)}{1/(pq^2)} = \frac{pq^2}{p^2r} = \frac{q^2}{pr}$$

Apply the AP condition.

The sequence is: $$3, \frac{3r}{pq}, \frac{3p^2}{q}, \frac{7q^2}{pr}$$ with common difference $$\frac{1}{2}$$.

From the first two terms: $$\frac{3r}{pq} - 3 = \frac{1}{2}$$, so $$\frac{3r}{pq} = \frac{7}{2}$$, giving $$r = \frac{7pq}{6}$$.

But $$r = pq + 1$$, so $$pq + 1 = \frac{7pq}{6}$$, giving $$6pq + 6 = 7pq$$, so $$pq = 6$$.

Therefore $$r = 6 + 1 = 7$$.

Find $$p$$ and $$q$$.

From the third term: $$\frac{3p^2}{q} = 3 + 2 \times \frac{1}{2} = 4$$, so $$3p^2 = 4q$$.

With $$pq = 6$$: $$q = 6/p$$. Substituting: $$3p^2 = 24/p$$, giving $$3p^3 = 24$$, so $$p^3 = 8$$, thus $$p = 2$$.

Then $$q = 6/2 = 3$$.

Verify with the fourth term.

$$\frac{7q^2}{pr} = \frac{7 \times 9}{2 \times 7} = \frac{63}{14} = \frac{9}{2} = 4.5$$.

Expected: $$3 + 3 \times 0.5 = 4.5$$. Confirmed!

Compute $$r - p - q$$.

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

The correct answer is Option 1: 2.