Let `veclamda=veca times (vecb +vecc), vecmu=vecb times (vecc+veca) and vecv=vecc times (veca+vecb)`, 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 given vectors and their relationships. We have:

1. \(\vec{\lambda} = \vec{a} \times (\vec{b} + \vec{c})\)
2. \(\vec{\nu} = \vec{b} \times (\vec{c} + \vec{a})\)
3. \(\vec{v} = \vec{c} \times (\vec{a} + \vec{b})\)

We need to explore the relationships between these vectors.

### Step 1: Expand the expressions for \(\vec{\lambda}\), \(\vec{\nu}\), and \(\vec{v}\)

Using the distributive property of the cross product, we can expand each vector:

\[ \vec{\lambda} = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \]

\[ \vec{\nu} = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} \]

\[ \vec{v} = \vec{c} \times \vec{a} + \vec{c} \times \vec{b} \]

### Step 2: Combine the vectors \(\vec{\lambda}\), \(\vec{\nu}\), and \(\vec{v}\)

Now, we add these three vectors together:

\[ \vec{\lambda} + \vec{\nu} + \vec{v} = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + (\vec{b} \times \vec{c} + \vec{b} \times \vec{a}) + (\vec{c} \times \vec{a} + \vec{c} \times \vec{b}) \]

### Step 3: Rearranging and applying properties of cross products

Using the property that \(\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})\), we can rewrite some terms:

\[ \vec{\lambda} + \vec{\nu} + \vec{v} = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} - \vec{a} \times \vec{b} + \vec{c} \times \vec{a} - \vec{b} \times \vec{c} \]

### Step 4: Simplifying the expression

Now, we can see that some terms will cancel out:

\[ \vec{\lambda} + \vec{\nu} + \vec{v} = \vec{a} \times \vec{c} + \vec{c} \times \vec{a} \]

Using the property \(\vec{c} \times \vec{a} = -(\vec{a} \times \vec{c})\):

\[ \vec{\lambda} + \vec{\nu} + \vec{v} = \vec{a} \times \vec{c} - \vec{a} \times \vec{c} = \vec{0} \]

### Conclusion

Since the sum of the vectors \(\vec{\lambda} + \vec{\nu} + \vec{v} = \vec{0}\), we conclude that the vectors are coplanar.