if `veca=hati+hatj+2hatk, vecb=hati+2hatj+2hatk` and `|vecc|=1`
Such that `[veca xx vecb vecb xx vecc vecc xx veca]` has maximum value, then the value of `|(veca xx vecb) xx vecc|^(2)` is (a) 0 (b) 1 (c) `4/3` (d) none of these

Options

A. 0
B. 1
C. `4/3`
D. none of these

Explanation

To solve the problem, we need to find the value of \(|(\vec{A} \times \vec{B}) \times \vec{C}|^2\) given that \(\vec{A} = \hat{i} + \hat{j} + 2\hat{k}\), \(\vec{B} = \hat{i} + 2\hat{j} + 2\hat{k}\), and \(|\vec{C}| = 1\) such that the expression \([\vec{A} \times \vec{B}, \vec{B} \times \vec{C}, \vec{C} \times \vec{A}]\) has a maximum value.

### Step 1: Calculate \(\vec{A} \times \vec{B}\)

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 1 & 2 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:

\[ = \hat{i} (1 \cdot 2 - 2 \cdot 2) - \hat{j} (1 \cdot 2 - 1 \cdot 2) + \hat{k} (1 \cdot 2 - 1 \cdot 1) \]

\[ = \hat{i} (2 - 4) - \hat{j} (2 - 2) + \hat{k} (2 - 1) \]

\[ = -2\hat{i} + 0\hat{j} + 1\hat{k} = -2\hat{i} + \hat{k} \]

### Step 2: Find the magnitude of \(\vec{A} \times \vec{B}\)

\[ |\vec{A} \times \vec{B}| = \sqrt{(-2)^2 + 0^2 + 1^2} = \sqrt{4 + 0 + 1} = \sqrt{5} \]

### Step 3: Determine \(\vec{C}\) for maximum value

For \([\vec{A} \times \vec{B}, \vec{B} \times \vec{C}, \vec{C} \times \vec{A}]\) to be maximum, \(\vec{C}\) should be perpendicular to both \(\vec{A}\) and \(\vec{B}\). Thus, we can take \(\vec{C}\) as a unit vector in the direction of \(\vec{A} \times \vec{B}\).

### Step 4: Calculate \(|(\vec{A} \times \vec{B}) \times \vec{C}|^2\)

Using the vector triple product identity:

\[ |(\vec{A} \times \vec{B}) \times \vec{C}|^2 = |\vec{A} \cdot \vec{C}|^2 |\vec{B}|^2 - |\vec{A} \cdot \vec{B}|^2 |\vec{C}|^2 \]

Since \(\vec{C}\) is perpendicular to both \(\vec{A}\) and \(\vec{B}\):

\[ \vec{A} \cdot \vec{C} = 0 \]

Thus:

\[ |(\vec{A} \times \vec{B}) \times \vec{C}|^2 = 0 \]

### Conclusion

The value of \(|(\vec{A} \times \vec{B}) \times \vec{C}|^2\) is \(0\).

### Final Answer

The correct option is (a) 0.