The imaginary part of $$(3 + 2\sqrt{-54})^{\frac{1}{2}} - (3 - 2\sqrt{-54})^{\frac{1}{2}}$$ can be:

1. Simplifying the Inner Terms

The expression involves $$\sqrt{-54}$$. We can write this in terms of $$i$$:

$$\sqrt{-54} = \sqrt{54} \cdot i = \sqrt{9 \times 6} \cdot i = 3\sqrt{6}i$$

Substituting this back into the expression, let:

$$z_1 = 3 + 2(3\sqrt{6}i) = 3 + 6\sqrt{6}i$$

$$z_2 = 3 - 2(3\sqrt{6}i) = 3 - 6\sqrt{6}i$$

The problem asks for the imaginary part of $$\sqrt{z_1} - \sqrt{z_2}$$.

2. Calculating the Square Root of a Complex Number

To find $$\sqrt{3 \pm 6\sqrt{6}i}$$, we look for a result in the form $$(x \pm iy)$$. We set:

$$(x + iy)^2 = 3 + 6\sqrt{6}i$$

$$x^2 - y^2 + 2xyi = 3 + 6\sqrt{6}i$$

Equating real and imaginary parts:

1. $$x^2 - y^2 = 3$$
2. $$2xy = 6\sqrt{6} \implies xy = 3\sqrt{6}$$

We also know the modulus identity $$|x+iy|^2 = \sqrt{3^2 + (6\sqrt{6})^2}$$:

$$x^2 + y^2 = \sqrt{9 + 216} = \sqrt{225} = 15$$

Now solve for $$x^2$$ and $$y^2$$:

$$(x^2 + y^2) + (x^2 - y^2) = 15 + 3 \implies 2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$$

$$(x^2 + y^2) - (x^2 - y^2) = 15 - 3 \implies 2y^2 = 12 \implies y^2 = 6 \implies y = \pm \sqrt{6}$$

Since $$xy$$ is positive ($$3\sqrt{6}$$), $$x$$ and $$y$$ must have the same sign:

$$\sqrt{z_1} = \pm(3 + \sqrt{6}i)$$

Similarly, for $$z_2$$, the imaginary part is negative, so $$x$$ and $$y$$ have opposite signs:

$$\sqrt{z_2} = \pm(3 - \sqrt{6}i)$$

3. Finding the Imaginary Part

Let $$E = \sqrt{z_1} - \sqrt{z_2}$$. Taking the primary roots (positive sign):

$$E = (3 + \sqrt{6}i) - (3 - \sqrt{6}i)$$

$$E = 3 + \sqrt{6}i - 3 + \sqrt{6}i$$

$$E = 2\sqrt{6}i$$

If we consider the different possible signs for square roots (specifically where $$\sqrt{z_1} = -(3 + \sqrt{6}i)$$ and $$\sqrt{z_2} = (3 - \sqrt{6}i)$$):

$$E = -3 - \sqrt{6}i - 3 + \sqrt{6}i = -6 \text{ (purely real)}$$

Or if $$\sqrt{z_1} = (3 + \sqrt{6}i)$$ and $$\sqrt{z_2} = -(3 - \sqrt{6}i)$$:

$$E = 3 + \sqrt{6}i + 3 - \sqrt{6}i = 6 \text{ (purely real)}$$

Or if both are negative:

$$E = (-3 - \sqrt{6}i) - (-3 + \sqrt{6}i) = -2\sqrt{6}i$$

The imaginary part of the result can be $$2\sqrt{6}$$ or $$-2\sqrt{6}$$. Based on the options provided:

$$\text{Im}(E) = -2\sqrt{6}$$

Final Answer:

The correct option is B, which is $$-2\sqrt{6}$$.