For three non - zero vectors `veca, vecb and vecc`, if `[(veca, vecb and vecc)]=4`, then `[vecaxx(vecb+2vecc)vecbxx(vecc-3veca)vecc xx(3veca+vecb)]`is equal to

Options

A. 12
B. 16
C. 84
D. 144

Answer

To solve the problem, we need to evaluate the expression given the scalar triple product of three vectors \( \vec{a}, \vec{b}, \vec{c} \) is equal to 4. The expression we need to evaluate is:

\[ [\vec{a} \times (\vec{b} + 2\vec{c}) \quad \vec{b} \times (\vec{c} - 3\vec{a}) \quad \vec{c} \times (3\vec{a} + \vec{b})] \]

### Step 1: Expand the Cross Products

We will first expand the cross products in the expression.

1. **First term**:
\[ \vec{a} \times (\vec{b} + 2\vec{c}) = \vec{a} \times \vec{b} + 2(\vec{a} \times \vec{c}) \]

2. **Second term**:
\[ \vec{b} \times (\vec{c} - 3\vec{a}) = \vec{b} \times \vec{c} - 3(\vec{b} \times \vec{a}) = \vec{b} \times \vec{c} + 3(\vec{a} \times \vec{b}) \quad \text{(using } \vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})\text{)} \]

3. **Third term**:
\[ \vec{c} \times (3\vec{a} + \vec{b}) = 3(\vec{c} \times \vec{a}) + \vec{c} \times \vec{b} \]

### Step 2: Combine the Terms

Now we can combine all the terms into a single expression:

\[ [\vec{a} \times \vec{b} + 2(\vec{a} \times \vec{c}) \quad \vec{b} \times \vec{c} + 3(\vec{a} \times \vec{b}) \quad 3(\vec{c} \times \vec{a}) + \vec{c} \times \vec{b}] \]

This simplifies to:

\[ [(\vec{a} \times \vec{b} + 3(\vec{a} \times \vec{b})) \quad (2(\vec{a} \times \vec{c})) \quad (3(\vec{c} \times \vec{a}) + \vec{c} \times \vec{b})] \]

### Step 3: Rearranging and Using Scalar Triple Product

Now we can rearrange the terms and factor out the scalar triple product:

\[ = [\vec{a} \times \vec{b} + 3\vec{a} \times \vec{b} \quad 2\vec{a} \times \vec{c} \quad 3\vec{c} \times \vec{a} + \vec{c} \times \vec{b}] \]

Using the property of the scalar triple product:

\[ [\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] \]

### Step 4: Calculate the Result

The scalar triple product is given as:

\[ [\vec{a}, \vec{b}, \vec{c}] = 4 \]

Thus, we can express the final determinant as:

\[ = 9 \cdot [\vec{a}, \vec{b}, \vec{c}]^2 = 9 \cdot (4)^2 = 9 \cdot 16 = 144 \]

### Final Answer

The value of the expression is:

\[ \boxed{144} \]