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 the A.P.: $$3, 3 \log_y x, 3 \log_z y, 7 \log_x z$$ with a common difference of $$d = \frac{1}{2}$$.

Let's evaluate each term by adding the common difference:

• Second term: $$3 \log_y x = 3 + \frac{1}{2} = \frac{7}{2} \implies \log_y x = \frac{7}{6}$$
• Third term: $$3 \log_z y = \frac{7}{2} + \frac{1}{2} = 4 \implies \log_z y = \frac{4}{3}$$
• Fourth term: $$7 \log_x z = 4 + \frac{1}{2} = \frac{9}{2} \implies \log_x z = \frac{9}{14}$$

From the exponential condition, let $$x^{pq^2} = y^{qr} = z^{p^2r} = k$$.

Taking logs, we can express the variables in terms of base $$k$$:

$$x = k^{\frac{1}{pq^2}}$$, $$y = k^{\frac{1}{qr}}$$, and $$z = k^{\frac{1}{p^2r}}$$.

Now, express the logarithms using these powers:

1. Use the first logarithm to find $$pq$$ and $$r$$:

$$\log_y x = \frac{\log x}{\log y} = \frac{\frac{1}{pq^2}}{\frac{1}{qr}} = \frac{r}{pq}$$

Equating this to our sequence value:

$$\frac{r}{pq} = \frac{7}{6}$$

We are given the relation $$r = pq + 1$$. Substitute this in:

$$\frac{pq + 1}{pq} = \frac{7}{6}$$

$$1 + \frac{1}{pq} = 1 + \frac{1}{6}$$

$$pq = 6$$

Since $$r = pq + 1$$, we get $$r = 7$$.

2. Use the second logarithm to separate $$p$$ and $$q$$:

$$\log_z y = \frac{\log y}{\log z} = \frac{\frac{1}{qr}}{\frac{1}{p^2r}} = \frac{p^2}{q}$$

Equating this to our sequence value:

$$\frac{p^2}{q} = \frac{4}{3}$$

Since we know $$pq = 6$$, we can write $$q = \frac{6}{p}$$. Substitute this in:

$$\frac{p^2}{\frac{6}{p}} = \frac{4}{3}$$

$$\frac{p^3}{6} = \frac{4}{3}$$

$$p^3 = 8$$

Since $$p$$ must be a positive integer, $$p = 2$$.

This means $$q = \frac{6}{2} = 3$$.

Finally, evaluate the required expression:

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