If `veca, vecb, vecc` are non coplanar non null vectors such that `[(veca, vecb, vecc)]=2` then `{[(vecaxxvecb, vecbxxvecc, veccxxveca)]}^(2)=`

Options

A. 4
B. 16
C. 8
D. none of these

Explanation

To solve the problem, we need to find the value of \({[(\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a)]}^2}\) given that \([(\vec{a}, \vec{b}, \vec{c})] = 2\).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We know that the scalar triple product \((\vec{a}, \vec{b}, \vec{c})\) is equal to 2. This can be interpreted as the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

2. **Using the Properties of Cross Products:**
We need to evaluate the expression \((\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a})\).
We can use the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} \]

3. **Calculating Each Cross Product:**
We can express the dot product of the cross products:
\[ (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})(\vec{b}) - (\vec{a} \cdot \vec{b})(\vec{c}) \]
Similarly, for the other terms:
\[ (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a})(\vec{c}) - (\vec{b} \cdot \vec{c})(\vec{a}) \]
\[ (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})(\vec{a}) - (\vec{c} \cdot \vec{a})(\vec{b}) \]

4. **Combining the Results:**
The entire expression can be simplified using the scalar triple product:
\[ \text{Let } V = [\vec{a}, \vec{b}, \vec{c}] = 2 \]
Then:
\[ (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = V^2 \]
Therefore:
\[ = (2)^2 = 4 \]

5. **Final Calculation:**
Now, we need to square this result:
\[ \{[(\vec{a} \times \vec{b}), (\vec{b} \times \vec{c}), (\vec{c} \times \vec{a})]\}^2 = 4^2 = 16 \]

### Final Answer:
\[ \{[(\vec{a} \times \vec{b}), (\vec{b} \times \vec{c}), (\vec{c} \times \vec{a})]\}^2 = 16 \]