$8, 12, 29$
Let the first term be $8$
$\therefore a = 8$ ...(i)
Let the $p^{th}$ ther be $12$
$\therefore ar^{p-1} = 12$ ...(ii)
and $q^{th}$ term be $27$
$\therefore ar^{q-1} = 27$ ...(iii)
(i) / (ii)
$\dfrac{a}{ar^{p-1}} = \dfrac{8}{12}$
$\Rightarrow r^{p-1} = \dfrac{3}{2} = 1.5$ ...(iv)
(iii) / (i)
$r^{q-1} = \dfrac{27}{8} = \left(\dfrac{3}{2}\right)^3$
$r^{q-1} = (1.5)^3$ ...(v)
(v) / (iv)
$\Rightarrow \dfrac{r^{q-1}}{r^{p-1}} = \dfrac{(1.5)^3}{1.5}$
$\Rightarrow r^{q-1-p+1} = (1.5)^2$
$\Rightarrow r^{q-p} = (1.5)^2$
$= (r^{p-1})^2$ ...from (iv)
$\Rightarrow r^{q-p} = r^{2p-2}$
$\therefore q-p = 2p-2$
$\Rightarrow \boxed{q=3p-2}$
for every distinct value of '$p$' there will be a district integer value of '$q$'
$\therefore $ Infinite no. of progression are possible.
option $C$