If `veca,vecb` and `vecc` are coplaner, then what is `(2vecaxx3vecb).4vecc+(5vecbxx3vecc).6veca` equal to?

Options

A. 114
B. 66
C. 0
D. `-66`

Answer

To solve the problem, we need to evaluate the expression \( (2\vec{a} \times 3\vec{b}) \cdot (4\vec{c}) + (5\vec{b} \times 3\vec{c}) \cdot (6\vec{a}) \) given that the vectors \(\vec{a}, \vec{b}, \vec{c}\) are coplanar.

### Step-by-Step Solution:

1. **Understanding Coplanarity**:
Since \(\vec{a}, \vec{b}, \vec{c}\) are coplanar, the scalar triple product \( \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \).

2. **Expanding the Expression**:
We can rewrite the expression:
\[ (2\vec{a} \times 3\vec{b}) \cdot (4\vec{c}) + (5\vec{b} \times 3\vec{c}) \cdot (6\vec{a}) \]
This simplifies to:
\[ 6(\vec{a} \times \vec{b}) \cdot (4\vec{c}) + 15(\vec{b} \times \vec{c}) \cdot (6\vec{a}) \]

3. **Factoring Out Constants**:
We can factor out the constants:
\[ 24(\vec{a} \times \vec{b}) \cdot \vec{c} + 90(\vec{b} \times \vec{c}) \cdot \vec{a} \]

4. **Using the Coplanarity Condition**:
Since \(\vec{a}, \vec{b}, \vec{c}\) are coplanar:
- \( (\vec{a} \times \vec{b}) \cdot \vec{c} = 0 \)
- \( (\vec{b} \times \vec{c}) \cdot \vec{a} = 0 \)

5. **Substituting Zero Values**:
Therefore, we substitute these values back into the expression:
\[ 24 \cdot 0 + 90 \cdot 0 = 0 \]

6. **Final Result**:
Thus, the final result of the expression is:
\[ 0 \]

### Conclusion:
The value of \( (2\vec{a} \times 3\vec{b}) \cdot (4\vec{c}) + (5\vec{b} \times 3\vec{c}) \cdot (6\vec{a}) \) is \( 0 \).