First, let's calculate the number of moles of NaOH that can be prepared from $$84\ \text{g}$$ of NaOH. The molar mass of NaOH is given as $$40\ \text{g/mol}$$.

The number of moles (n) is calculated using the formula:

$$ n = \frac{mass}{molar\ mass} $$

So for our case:

$$ n = \frac{84\ \text{g}}{40\ \text{g/mol}} = 2.1\ \text{mol} $$

Now that we know the number of moles, we can find out the volume of a $$3\ \text{M}$$ NaOH solution that can be prepared from it. The concentration (C) of a solution is related to the number of moles (n) and volume (V) by the following formula:

$$ C = \frac{n}{V} $$

Where: C = concentration in molarity (M) n = number of moles V = volume in liters (L) - Note that $$1\ \text{dm}^3 = 1\ \text{L}$$.

Since we want to find the volume (V), we can rearrange the formula to solve for V:

$$ V = \frac{n}{C} $$

Using the moles of NaOH and the concentration for the preparation:

$$ V = \frac{2.1\ \text{mol}}{3\ \text{M}} $$

Calculate the volume:

$$ V = \frac{2.1}{3} = 0.7\ \text{L} $$

To convert this volume to $$\text{dm}^3$$ (which is equivalent to liters), we use the conversion factor $$1\ \text{L} = 1\ \text{dm}^3$$. Therefore:

$$ V = 0.7\ \text{dm}^3 $$

To express this volume as $$\times 10^{-1}\ \text{dm}^3$$, we can write:

$$ V = 7 \times 10^{-1}\ \text{dm}^3 $$

Therefore, the volume of $$3\ \text{M}$$ NaOH solution which can be prepared from $$84\ \text{g}$$ of NaOH is $$7 \times 10^{-1}\ \text{dm}^3$$.