Given `vec(a) = xhat(i) + yhat(j) + 2hat(k) , vec(b) = hat(i) - hat(j) + hat(k), hat(c) = hat(i) + 2hat(j) , (vec(a) wedge vec(b)) = pi//2 , vec(a).vec(c) = 4` then

Options

A. `[vec(a)vec(b)vec(c)]^(2) = |vec(a)|`
B. `[vec(a) vec(b) vec(c)]=|vec(a)|`
C. `[vec(a) vec(b) vec(c)] = 0`
D. `[vec(a) vec(b) vec(c)] =|vec(a)|^(2)`

Answer

To solve the given problem, we need to analyze the vectors and their relationships based on the conditions provided. Let's break down the solution step by step.

### Given Vectors:
1. \(\vec{a} = x \hat{i} + y \hat{j} + 2 \hat{k}\)
2. \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\)
3. \(\vec{c} = \hat{i} + 2 \hat{j}\)

### Conditions:
1. \(\vec{a} \wedge \vec{b} = \frac{\pi}{2}\) (This means \(\vec{a}\) and \(\vec{b}\) are perpendicular)
2. \(\vec{a} \cdot \vec{c} = 4\)

### Step 1: Use the Perpendicular Condition
Since \(\vec{a}\) and \(\vec{b}\) are perpendicular, their dot product is zero:
\[ \vec{a} \cdot \vec{b} = 0 \]
Calculating the dot product:
\[ (x \hat{i} + y \hat{j} + 2 \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = x(1) + y(-1) + 2(1) = x - y + 2 = 0 \]
This gives us our first equation:
\[ x - y + 2 = 0 \quad \text{(1)} \]

### Step 2: Use the Second Condition
Now, using the second condition:
\[ \vec{a} \cdot \vec{c} = 4 \]
Calculating the dot product:
\[ (x \hat{i} + y \hat{j} + 2 \hat{k}) \cdot (\hat{i} + 2 \hat{j}) = x(1) + y(2) + 2(0) = x + 2y = 4 \]
This gives us our second equation:
\[ x + 2y = 4 \quad \text{(2)} \]

### Step 3: Solve the System of Equations
Now we have a system of equations:
1. \(x - y + 2 = 0\)
2. \(x + 2y = 4\)

From equation (1), we can express \(x\) in terms of \(y\):
\[ x = y - 2 \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ (y - 2) + 2y = 4 \]
\[ 3y - 2 = 4 \]
\[ 3y = 6 \implies y = 2 \]

Now substituting \(y = 2\) back into equation (3):
\[ x = 2 - 2 = 0 \]

### Step 4: Find the Vectors
Now we have:
\[ x = 0, \quad y = 2 \]
Thus, the vector \(\vec{a}\) becomes:
\[ \vec{a} = 0 \hat{i} + 2 \hat{j} + 2 \hat{k} = 2 \hat{j} + 2 \hat{k} \]

### Step 5: Calculate the Magnitude of \(\vec{a}\)
The magnitude of \(\vec{a}\) is:
\[ |\vec{a}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{0 + 4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 6: Calculate the Scalar Triple Product
The scalar triple product \(\text{box}(\vec{a}, \vec{b}, \vec{c})\) can be calculated using the determinant:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = \begin{vmatrix} 0 & 2 & 2 \\ 1 & -1 & 1 \\ 1 & 2 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = 0 \cdot \begin{vmatrix} -1 & 1 \\ 2 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} \]
\[ = 0 - 2(1 \cdot 0 - 1 \cdot 1) + 2(1 \cdot 2 - (-1) \cdot 1) \]
\[ = 0 + 2(0 - 1) + 2(2 + 1) = 0 + (-2) + 6 = 4 \]

### Conclusion
Thus, we find that:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = 4 \]
And since \( |\vec{a}|^2 = 8 \), we can conclude:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = |\vec{a}|^2 \]

### Final Answer
The answer is option D.