\(\mathsf{the\:three\:terms\:GP\:are\:\dfrac{a}{R},a,ar}\)
\(\mathsf{product\:of\:three\:numbers\:are\:a^3=27}\)
→ \(\mathsf{a=3}\)............................(1)
\(\mathsf{substitute\: in\:1} \)
→\(\mathsf{\dfrac{3}{r}+3r+3=S}\)
\(\mathsf{here\:r>0}\)\
→\(\mathsf{\dfrac{\dfrac{3}{r}+3r}{2}}>\sqrt{3^2}\)
→\(\mathsf{\dfrac{3}{r}+3r \geq 6}\)............................(2)
→\(\mathsf{for\: r<0}\)
\(\mathsf{\dfrac{3}{r}+3r \leq -6}\).........................(3)
\(\mathsf{from\:2\:and\:3\:we\:get}\)
so option is 4