Let `veca = 2hati + hatj + hatk, and vecb = hati+ hatj ` if c is a vector such that `veca .vecc = |vecc|, |vecc -veca| = 2sqrt2` and the angle between `veca xx vecb and vec is 30^(@)` , then `|(veca xx vecb)|xx vecc|` is equal to

Options

A. `2//3`
B. `3//2`
C. 2
D. 3

Explanation

To solve the problem step by step, we will follow the given conditions and use vector operations accordingly.

### Step 1: Define the vectors
We have the vectors:
\[ \vec{a} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} + \hat{j} \]

### Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\)
To find \(\vec{a} \times \vec{b}\), we set up the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot 0 - 1 \cdot 1) - \hat{j} (2 \cdot 0 - 1 \cdot 1) + \hat{k} (2 \cdot 1 - 1 \cdot 1) \]
\[ = \hat{i} (-1) - \hat{j} (-1) + \hat{k} (2 - 1) \]
\[ = -\hat{i} + \hat{j} + \hat{k} \]

Thus,
\[ \vec{a} \times \vec{b} = -\hat{i} + \hat{j} + \hat{k} \]

### Step 3: Calculate the magnitude of \(\vec{a} \times \vec{b}\)
Now, we find the magnitude:
\[ |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 4: Use the given conditions about vector \(\vec{c}\)
We know:
1. \(\vec{a} \cdot \vec{c} = |\vec{c}|\)
2. \(|\vec{c} - \vec{a}| = 2\sqrt{2}\)

From the second condition, we square both sides:
\[ |\vec{c} - \vec{a}|^2 = 8 \]
This expands to:
\[ |\vec{c}|^2 - 2\vec{c} \cdot \vec{a} + |\vec{a}|^2 = 8 \]

### Step 5: Calculate \(|\vec{a}|\)
\[ |\vec{a}| = \sqrt{(2)^2 + (1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 6: Substitute known values
Using \(\vec{a} \cdot \vec{c} = |\vec{c}|\), we can replace \(\vec{c} \cdot \vec{a}\) with \(|\vec{c}|\):
\[ |\vec{c}|^2 - 2|\vec{c}| + 6 = 8 \]
This simplifies to:
\[ |\vec{c}|^2 - 2|\vec{c}| - 2 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ |\vec{c}| = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \]
Since magnitudes are positive, we take:
\[ |\vec{c}| = 1 + \sqrt{3} \]

### Step 8: Find \(|(\vec{a} \times \vec{b}) \times \vec{c}|\)
Using the formula:
\[ |(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin \theta \]
where \(\theta = 30^\circ\) and \(\sin 30^\circ = \frac{1}{2}\):
\[ |(\vec{a} \times \vec{b}) \times \vec{c}| = \sqrt{3} \cdot (1 + \sqrt{3}) \cdot \frac{1}{2} \]
\[ = \frac{\sqrt{3} (1 + \sqrt{3})}{2} = \frac{\sqrt{3} + 3}{2} \]

### Final Answer
Thus, the final answer is:
\[ \boxed{\frac{\sqrt{3} + 3}{2}} \]