If a, b, c are the pth, qth and rth terms of a G.P., then the angle between the vector $\vec u =(\log a) \hat i + (\log b)\hat j + (\log c)\hat k$ and $\vec v = (q-r)\hat i + (r-p)\hat j+(p −q)\hat k$, is

$\frac{π}{2}$

Let A be the first term and R be the common ratio of the given GP. Then,

$a = AR^{p-1}, b = AR^{q-1}, c = AR^{r-1}$

$⇒a^{q-r}b^{r-P} c^{p-q}=A^0R^0=1$

$⇒(q-r)\log a +(r-p)\log b + (p-q)\log c = 0$

$⇒\vec u.\vec v=0⇒\vec u⊥\vec v$

Hence, require angle is $π/2$.