Step 1 : Understanding the Question:
The question explores a high-level identity within vector algebra involving the Scalar Triple Product (STP). We are given the STP of three primary vectors $\vec{a}, \vec{b}$, and $\vec{c}$, which geometrically represents the volume of a parallelepiped. We are then asked to find the STP of a new set of three vectors, where each new vector is the cross product of two of the original vectors. This problem tests the knowledge of vector identities and the relationships between primary and reciprocal-like vector sets.
Step 2 : Key Formulas and approach:
The Scalar Triple Product of three vectors $\vec{u}, \vec{v}, \vec{w}$ is $[\vec{u} \ \vec{v} \ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$. A very important property of the STP is that the product of the cross products $(\vec{a} \times \vec{b})$, $(\vec{b} \times \vec{c})$, and $(\vec{c} \times \vec{a})$ relates directly to the square of the original STP.
The fundamental identity used here is:
\[ [\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = [\vec{a} \ \vec{b} \ \vec{c}]^2 \]
Our approach is to apply this identity directly using the provided numerical value.
Step 3 : Detailed Explanation:
Start by identifying the given value: $[\vec{a} \ \vec{b} \ \vec{c}] = 3$.
Recall the derivation logic of the identity: Let $\vec{X} = \vec{b} \times \vec{c}$. Then the cross product $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})$ involves a vector triple product which simplifies using the property $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$.
Specifically, $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [(\vec{b} \times \vec{c}) \cdot \vec{a}]\vec{c} - [(\vec{b} \times \vec{c}) \cdot \vec{c}]\vec{a}$.
Since a cross product is perpendicular to its constituents, $(\vec{b} \times \vec{c}) \cdot \vec{c} = 0$, reducing the expression to $[\vec{a} \ \vec{b} \ \vec{c}]\vec{c}$.
Now, the whole triple product becomes: $(\vec{a} \times \vec{b}) \cdot ([\vec{a} \ \vec{b} \ \vec{c}]\vec{c}) = [\vec{a} \ \vec{b} \ \vec{c}] \cdot (\vec{a} \times \vec{b} \cdot \vec{c})$.
This is equivalent to $[\vec{a} \ \vec{b} \ \vec{c}] \cdot [\vec{a} \ \vec{b} \ \vec{c}] = [\vec{a} \ \vec{b} \ \vec{c}]^2$.
Substituting the given value: $3^2 = 9$.
Step 4 : Final Answer:
The value of the scalar triple product of the cross products is 9.