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

We are given $$x^{pq^2} = y^{qr} = z^{p^2r}$$ with $$r = pq + 1$$, and the four quantities $$3, 3\log_y x, 3\log_z y, 7\log_x z$$ form an arithmetic progression with common difference $$\frac{1}{2}$$.

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

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

Hence the four terms of the progression are $$3,\;\frac{3r}{pq},\;\frac{3p^2}{q},\;\frac{7q^2}{pr},$$ and the common difference is $$\tfrac12\,. $$

Equating the second term to $$3 + \tfrac12 = \tfrac72$$ gives $$\frac{3r}{pq} = \frac{7}{2},\quad 6r = 7pq,\quad r = \frac{7pq}{6}.$$ Since also $$r = pq + 1,$$ we obtain $$\frac{7pq}{6} = pq + 1\;\Longrightarrow\;\frac{pq}{6} = 1\;\Longrightarrow\;pq = 6,\;r = 7.$$

Equating the third term to $$3 + 2\times\tfrac12 = 4$$ yields $$\frac{3p^2}{q} = 4\;\Longrightarrow\;3p^2 = 4q.$$ With $$q = \tfrac{6}{p}$$ this gives $$3p^2 = \tfrac{24}{p}\;\Longrightarrow\;p^3 = 8\;\Longrightarrow\;p = 2,\;q = 3.$$

The fourth term is $$\frac{7q^2}{pr} = \frac{7\times 9}{2\times 7} = \frac{9}{2} = 4.5 = 3 + 3\times\tfrac12,$$ which confirms the common difference.

Finally, $$r - p - q = 7 - 2 - 3 = 2.$$ The answer is Option 1: 2.