1. Let the first three terms of the G.P. be $\frac{a}{r},\ a,\ ar$ where $a$ is the middle term.

2. Use the product condition Their product is $\frac{a}{r}\cdot a \cdot ar = a^3$ Given this product is $27$, so $a^3=27 \implies a=3$

3. Write the sum of the three terms $S=\frac{3}{r}+3+3r=3\left(r+1+\frac{1}{r}\right), \qquad r\neq 0$

4. Let $x=r+\frac{1}{r}$ Then $S=3(x+1)$

5. Find the possible values of $x$ For real $r\neq 0$, $r+\frac{1}{r} \ge 2 \quad \text{or} \quad r+\frac{1}{r} \le -2$ Hence, $x\in (-\infty,-2]\cup[2,\infty)$

6. Transform this to the range of $S$ Since $S=3(x+1)$ we get:

   • if $x\le -2$, then $S\le 3(-2+1)=-3$
   • if $x\ge 2$, then $S\ge 3(2+1)=9$

   Therefore, $S\in (-\infty,-3]\cup[9,\infty)$

7. Match with the options This is exactly Option D.

Verification with stored answer: Stored correct answer is D, which matches our derived answer.