Let $$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j}$$. If $$\vec{c}$$ is a vector such that $$\vec{a} \cdot \vec{c} = |\vec{c}|$$, $$|\vec{c} - \vec{a}| = 2\sqrt{2}$$ and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is 30°, then $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ equals:

Given vectors are $$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j}$$. We need to find $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ where $$\vec{c}$$ satisfies three conditions: $$\vec{a} \cdot \vec{c} = |\vec{c}|$$, $$|\vec{c} - \vec{a}| = 2\sqrt{2}$$, and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is 30°.

First, compute $$\vec{a} \times \vec{b}$$. The cross product is found using the determinant formula:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \\ \end{vmatrix}$$

Expanding along the first row:

$$\hat{i} \left( (1)(0) - (-2)(1) \right) - \hat{j} \left( (2)(0) - (-2)(1) \right) + \hat{k} \left( (2)(1) - (1)(1) \right)$$

Calculate each component:

For $$\hat{i}$$: $$1 \cdot 0 - (-2) \cdot 1 = 0 + 2 = 2$$, so $$2\hat{i}$$.

For $$\hat{j}$$: minus sign, then $$2 \cdot 0 - (-2) \cdot 1 = 0 + 2 = 2$$, so $$-2\hat{j}$$.

For $$\hat{k}$$: $$2 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1$$, so $$1\hat{k}$$.

Thus, $$\vec{a} \times \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$$.

Now, find its magnitude:

$$|\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3.$$

Denote $$\vec{d} = \vec{a} \times \vec{b}$$, so $$|\vec{d}| = 3$$.

The magnitude $$|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{d} \times \vec{c}|$$. The magnitude of the cross product is given by $$|\vec{d} \times \vec{c}| = |\vec{d}| |\vec{c}| \sin \theta$$, where $$\theta$$ is the angle between $$\vec{d}$$ and $$\vec{c}$$. Given $$\theta = 30^\circ$$, $$\sin 30^\circ = \frac{1}{2}$$, so:

$$|\vec{d} \times \vec{c}| = 3 \cdot |\vec{c}| \cdot \frac{1}{2} = \frac{3}{2} |\vec{c}|.$$

To find $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$, we need $$|\vec{c}|$$. Use the conditions $$\vec{a} \cdot \vec{c} = |\vec{c}|$$ and $$|\vec{c} - \vec{a}| = 2\sqrt{2}$$.

Let $$|\vec{c}| = c$$, so $$\vec{a} \cdot \vec{c} = c$$.

Compute $$|\vec{a}|^2$$:

$$|\vec{a}|^2 = (2)^2 + (1)^2 + (-2)^2 = 4 + 1 + 4 = 9.$$

Now, $$|\vec{c} - \vec{a}| = 2\sqrt{2}$$, so $$|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$$.

Expand the magnitude squared:

$$|\vec{c} - \vec{a}|^2 = (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = \vec{c} \cdot \vec{c} - 2 \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{a} = |\vec{c}|^2 - 2 (\vec{a} \cdot \vec{c}) + |\vec{a}|^2.$$

Substitute known values:

$$c^2 - 2(c) + 9 = 8.$$

Simplify:

$$c^2 - 2c + 9 - 8 = 0 \implies c^2 - 2c + 1 = 0.$$

Factor:

$$(c - 1)^2 = 0 \implies c = 1.$$

Thus, $$|\vec{c}| = 1$$.

Now substitute back:

$$|\vec{d} \times \vec{c}| = \frac{3}{2} \cdot 1 = \frac{3}{2}.$$

Therefore, $$|(\vec{a} \times \vec{b}) \times \vec{c}| = \frac{3}{2}$$.

Hence, the correct answer is Option D.