Given, $$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j}$$. Let $$\vec{c}$$ be a vector such that $$|\vec{c} - \vec{a}| = 3$$, $$|\vec{a} \times \vec{b} \times \vec{c}| = 3$$ and the angle between $$\vec{c}$$ and $$\vec{a} \times \vec{b}$$ be 30°. Then $$\vec{a} \cdot \vec{c}$$ is equal to:

We have $$\vec a = 2\hat i + \hat j - 2\hat k$$ and $$\vec b = \hat i + \hat j$$. Let us first evaluate the cross product $$\vec a \times \vec b$$ because it will appear again and again.

Using the determinant form of the cross product,

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & -2\\ 1 & 1 & 0 \end{vmatrix} = \hat i\,(1\cdot0-(-2)\cdot1)\;-\;\hat j\,(2\cdot0-(-2)\cdot1)\;+\;\hat k\,(2\cdot1-1\cdot1)$$

$$\qquad\;= \hat i\,(0+2)\;-\;\hat j\,(0+2)\;+\;\hat k\,(2-1) = 2\hat i - 2\hat j + \hat k.$$

So we denote $$\vec d = \vec a \times \vec b = 2\hat i - 2\hat j + \hat k.$$

The magnitude of $$\vec d$$ is

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

According to the question, the angle between $$\vec c$$ and $$\vec d$$ is $$30^\circ$$. We also know the magnitude of the cross product of these two vectors: $$|\vec d \times \vec c| = 3.$$ For any two vectors $$\vec p$$ and $$\vec q$$ the formula for the magnitude of their cross product is

$$|\vec p \times \vec q| = |\vec p|\,|\vec q| \sin\theta,$$

where $$\theta$$ is the angle between them. Applying this with $$\vec p=\vec d$$, $$\vec q=\vec c$$ and $$\theta = 30^\circ$$ gives

$$|\vec d \times \vec c| = |\vec d|\,|\vec c| \sin 30^\circ = 3 \, |\vec c| \left(\frac12\right) = \frac{3}{2}\,|\vec c|.$$

The question states that this value equals 3, so

$$\frac{3}{2}\,|\vec c| = 3 \;\;\Longrightarrow\;\; |\vec c| = \frac{3 \times 2}{3} = 2.$$

Thus $$|\vec c| = 2.$$ Another piece of information provided is $$|\vec c - \vec a| = 3.$$ To use this we expand the square of the magnitude. The standard identity is

$$|\vec u - \vec v|^2 = |\vec u|^2 + |\vec v|^2 - 2\,\vec u\cdot\vec v.$$

Putting $$\vec u = \vec c$$ and $$\vec v = \vec a,$$ we write

$$|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\,\vec a\cdot\vec c.$$

We already have $$|\vec c - \vec a| = 3,$$ so the left-hand side is $$3^2 = 9.$$ We have also just found $$|\vec c| = 2,$$ therefore $$|\vec c|^2 = 4.$$ Next we need $$|\vec a|.$$ Its magnitude is

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

Substituting these values into the identity,

$$9 = 4 + 9 - 2\,\vec a\cdot\vec c.$$

Simplifying the right-hand side, $$4 + 9 = 13,$$ hence

$$9 = 13 - 2\,\vec a\cdot\vec c.$$

Now shift terms to isolate the dot product:

$$-2\,\vec a\cdot\vec c = 9 - 13 = -4,$$

so

$$\vec a\cdot\vec c = \frac{-4}{-2} = 2.$$

Therefore $$\vec a \cdot \vec c = 2.$$

Hence, the correct answer is Option B.