Let `barlamda=bara times (barb +barc), barmu=barb times (barc+bara) and barv=barc times (bara+barb)`, Then

Options

A. `barlamda+barmu=barv`
B. `barlamda,barmu and barv` are coplanar
C. `barlamda+barmu+barv=0`
D. `barlamda+barv=barmu`

Answer

To solve the problem, we need to analyze the expressions given for \(\lambda\), \(\mu\), and \(\nu\):

1. **Given Expressions**:
- \(\lambda = \mathbf{a} \times (\mathbf{b} + \mathbf{c})\)
- \(\mu = \mathbf{b} \times (\mathbf{c} + \mathbf{a})\)
- \(\nu = \mathbf{c} \times (\mathbf{a} + \mathbf{b})\)

2. **Expanding Each Expression**:
- For \(\lambda\):
\[ \lambda = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} \]
- For \(\mu\):
\[ \mu = \mathbf{b} \times \mathbf{c} + \mathbf{b} \times \mathbf{a} \]
- For \(\nu\):
\[ \nu = \mathbf{c} \times \mathbf{a} + \mathbf{c} \times \mathbf{b} \]

3. **Adding the Three Expressions**:
\[ \lambda + \mu + \nu = (\mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}) + (\mathbf{b} \times \mathbf{c} + \mathbf{b} \times \mathbf{a}) + (\mathbf{c} \times \mathbf{a} + \mathbf{c} \times \mathbf{b}) \]

4. **Rearranging the Terms**:
\[ \lambda + \mu + \nu = \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{a} + \mathbf{a} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{b} \]

5. **Using the Property of Cross Products**:
- Recall that \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\) and similar for others:
\[ \lambda + \mu + \nu = \mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} - \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} - \mathbf{c} \times \mathbf{b} \]
This simplifies to:
\[ \lambda + \mu + \nu = 0 \]

6. **Conclusion**:
- Therefore, we conclude that:
\[ \lambda + \mu + \nu = 0 \]