Let $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$, $$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$$, $$\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$$ and $$\vec{d}$$ be a vector such that $$\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$$ and $$\vec{a} \cdot \vec{d} = 4$$. Then $$|\vec{a} \times \vec{d}|^2$$ is equal to ________.

Given $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$, $$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$$, $$\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$$.

We need $$\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$$, which gives $$(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$$.

$$\vec{b} - \vec{c} = (3-2)\hat{i} + (-3+1)\hat{j} + (3-2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$$

Since $$(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$$, $$\vec{d}$$ is parallel to $$\vec{b} - \vec{c} = \hat{i} - 2\hat{j} + \hat{k}$$.

So $$\vec{d} = t(\hat{i} - 2\hat{j} + \hat{k})$$ for some scalar $$t$$.

Using $$\vec{a} \cdot \vec{d} = 4$$: $$t(1 - 4 + 1) = 4$$, so $$t(-2) = 4$$, giving $$t = -2$$.

Therefore, $$\vec{d} = -2\hat{i} + 4\hat{j} - 2\hat{k}$$.

Now, $$\vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix}$$

$$= \hat{i}(-4 - 4) - \hat{j}(-2 + 2) + \hat{k}(4 + 4) = -8\hat{i} + 0\hat{j} + 8\hat{k}$$

$$|\vec{a} \times \vec{d}|^2 = 64 + 0 + 64 = 128$$

Hence, the correct answer is 128.