Let $$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j}$$. 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 $$\frac{\pi}{6}$$, then the value of $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ is:

We are given $$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j}$$. First, we compute $$\vec{a} \times \vec{b}$$:

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

So $$|\vec{a} \times \vec{b}| = \sqrt{4 + 4 + 1} = 3$$. Also $$|\vec{a}| = \sqrt{4 + 1 + 4} = 3$$.

From the condition $$|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$$, we expand to get $$|\vec{c}|^2 - 2\vec{a} \cdot \vec{c} + |\vec{a}|^2 = 8$$. Using $$\vec{a} \cdot \vec{c} = |\vec{c}|$$ and $$|\vec{a}|^2 = 9$$:

$$|\vec{c}|^2 - 2|\vec{c}| + 9 = 8 \implies |\vec{c}|^2 - 2|\vec{c}| + 1 = 0 \implies (|\vec{c}| - 1)^2 = 0$$

Thus $$|\vec{c}| = 1$$. Now we use the identity $$|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}||\vec{c}|\sin\theta$$, where $$\theta = \frac{\pi}{6}$$ is the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$:

$$|(\vec{a} \times \vec{b}) \times \vec{c}| = 3 \cdot 1 \cdot \sin\frac{\pi}{6} = 3 \cdot \frac{1}{2} = \frac{3}{2}$$