If three successive terms of a G.P. with common ratio $$r$$ $$(r > 1)$$ are the length of the sides of a triangle and $$\lfloor r \rfloor$$ denotes the greatest integer less than or equal to r, then $$3\lfloor r \rfloor + \lfloor -r \rfloor$$ is equal to:

Let the three successive terms of the G.P. be $$a,\;ar,\;ar^{2}$$ with common ratio $$r$$, where $$r \gt 1$$.

For these three numbers to be the lengths of the sides of a triangle, they must satisfy the triangle inequality.
Since $$r \gt 1$$, the terms are in increasing order: $$a \lt ar \lt ar^{2}$$. The only non-trivial inequality is

$$a + ar \gt ar^{2}\;.$$

Dividing by the positive number $$a$$ gives

$$1 + r \gt r^{2}\;.$$

Re-arranging,

$$r^{2} - r - 1 \lt 0 \;.$$

The roots of $$r^{2} - r - 1 = 0$$ are $$r = \dfrac{1 \pm \sqrt{5}}{2}$$.
Let $$\alpha = \dfrac{1 + \sqrt{5}}{2} \approx 1.618\;.$$

Because the parabola opens upward, $$r^{2} - r - 1 \lt 0$$ for $$-\;0.618\;\lt r \lt \alpha$$.
With the given condition $$r \gt 1$$, we obtain

$$1 \lt r \lt \alpha\;.$$

Thus $$r$$ lies strictly between $$1$$ and $$\alpha \;( \lt 2)$$. Therefore
$$\lfloor r \rfloor = 1\;.$$

Next, $$-r$$ lies between $$-\,\alpha$$ and $$-1$$, i.e. $$-1.618\lt -r\lt -1$$, so

$$\lfloor -r \rfloor = -2\;.$$

Hence

$$3\lfloor r \rfloor + \lfloor -r \rfloor \;=\; 3(1) + (-2) \;=\; 1\;.$$

Therefore, the required value is $$1$$.