The sum of the first three terms of G.P. is $$S$$ and their product is 27. Then all such $$S$$ lie in:

Let the three consecutive terms of the given geometric progression be $$a,\;ar,\;ar^{2}$$ where $$a$$ is the first term and $$r$$ (with $$r \neq 0$$) is the common ratio.

We are told that the product of these three terms is $$27$$. Writing this information in algebraic form we have

$$a \cdot ar \cdot ar^{2}=27.$$

Simplifying the left‐hand side gives $$a^{3}r^{3}=27,$$ and recognising that $$a^{3}r^{3}=(ar)^{3},$$ we obtain

$$(ar)^{3}=27.$$

Taking the real cube root on both sides we get

$$ar=3.$$

This relation allows us to express $$a$$ in terms of $$r$$, namely

$$a=\dfrac{3}{r}.$$

Next, let us translate the statement about the sum of the first three terms. The sum is given to be $$S$$, so

$$S=a+ar+ar^{2}.$$

Substituting $$a=\dfrac{3}{r}$$ into the above expression yields

$$S=\dfrac{3}{r}+\dfrac{3}{r}\,r+\dfrac{3}{r}\,r^{2}.$$

Carrying out the multiplications inside the terms we obtain

$$S=\dfrac{3}{r}+3+3r.$$

It is convenient to factor out the common factor $$3$$:

$$S=3\!\left(\dfrac{1}{r}+1+r\right).$$

We now introduce the standard substitution $$t=r+\dfrac{1}{r},$$ which frequently appears when dealing with expressions containing both $$r$$ and $$\dfrac{1}{r}$$. With this substitution we can rewrite the bracketed term as follows:

$$\dfrac{1}{r}+1+r=\Bigl(r+\dfrac{1}{r}\Bigr)+1=t+1.$$

Hence the sum $$S$$ becomes

$$S=3(t+1)=3t+3.$$

All that remains is to determine the range of the variable $$t=r+\dfrac{1}{r}$$ for real, non-zero $$r$$. The well-known inequality based on the AM-GM (or by direct calculus) states that for any real $$r\neq 0$$,

$$r+\dfrac{1}{r}\;\ge\;2 \quad\text{or}\quad r+\dfrac{1}{r}\;\le\;-2.$$

In interval notation, $$t \in (-\infty,-2] \cup [2,\infty).$$

Since $$S=3t+3$$ is a linear transformation of $$t$$, its range is obtained by applying the same transformation to each part of the above interval:

For $$t \le -2$$: $$S=3t+3 \le 3(-2)+3 = -6+3 = -3,$$ giving the interval $$(-\infty,-3].$$

For $$t \ge 2$$: $$S=3t+3 \ge 3(2)+3 = 6+3 = 9,$$ giving the interval $$[9,\infty).$$

Together, the possible values of $$S$$ lie in

$$(-\infty,-3]\;\cup\;[9,\infty).$$

Comparing this result with the options provided, we see that it matches Option C.

Hence, the correct answer is Option C.