If `vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)`
Such that `x+y+z!=0` and `vecr.(veca+vecb+vecc)=x+y+z`, then `[veca vecb vecc]=`

Options

A. 0
B. 1
C. `-1`
D. 2

Explanation

To solve the problem, we need to find the determinant \([ \vec{a} \, \vec{b} \, \vec{c} ]\) given the equation:

\[ \vec{r} = x (\vec{a} \times \vec{b}) + y (\vec{b} \times \vec{c}) + z (\vec{c} \times \vec{a}) \]

and the condition:

\[ \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) = x + y + z \]

### Step 1: Substitute \(\vec{r}\) into the dot product equation

We start by substituting the expression for \(\vec{r}\) into the dot product equation:

\[ (x (\vec{a} \times \vec{b}) + y (\vec{b} \times \vec{c}) + z (\vec{c} \times \vec{a})) \cdot (\vec{a} + \vec{b} + \vec{c}) = x + y + z \]

### Step 2: Expand the left-hand side

Now we will expand the left-hand side using the distributive property of the dot product:

\[ x ((\vec{a} \times \vec{b}) \cdot (\vec{a} + \vec{b} + \vec{c})) + y ((\vec{b} \times \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})) + z ((\vec{c} \times \vec{a}) \cdot (\vec{a} + \vec{b} + \vec{c})) \]

### Step 3: Evaluate each dot product

Using the property of the cross product, we know that:

- \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\)
- \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\)

Thus, we only need to consider the term with \(\vec{c}\):

\[ x (\vec{a} \times \vec{b}) \cdot \vec{c} + y ((\vec{b} \times \vec{c}) \cdot \vec{a}) + y ((\vec{b} \times \vec{c}) \cdot \vec{b}) + z ((\vec{c} \times \vec{a}) \cdot \vec{a}) + z ((\vec{c} \times \vec{a}) \cdot \vec{b}) + z ((\vec{c} \times \vec{a}) \cdot \vec{c}) \]

The terms involving \(\vec{a}\) and \(\vec{b}\) will be zero, leaving us with:

\[ x (\vec{a} \times \vec{b}) \cdot \vec{c} + y (\vec{b} \times \vec{c}) \cdot \vec{a} + z (\vec{c} \times \vec{a}) \cdot \vec{b} \]

### Step 4: Rewrite the dot products as determinants

Using the identity \((\vec{u} \times \vec{v}) \cdot \vec{w} = [\vec{u} \, \vec{v} \, \vec{w}]\), we can rewrite our expression:

\[ x [\vec{a} \, \vec{b} \, \vec{c}] + y [\vec{b} \, \vec{c} \, \vec{a}] + z [\vec{c} \, \vec{a} \, \vec{b}] = x + y + z \]

### Step 5: Factor out the determinant

Notice that the determinants can be rearranged without changing their sign:

\[ x [\vec{a} \, \vec{b} \, \vec{c}] + y [\vec{a} \, \vec{b} \, \vec{c}] + z [\vec{a} \, \vec{b} \, \vec{c}] = (x + y + z) [\vec{a} \, \vec{b} \, \vec{c}] \]

### Step 6: Equate and solve for the determinant

Now we have:

\[ (x + y + z) [\vec{a} \, \vec{b} \, \vec{c}] = x + y + z \]

Since \(x + y + z \neq 0\), we can divide both sides by \(x + y + z\):

\[ [\vec{a} \, \vec{b} \, \vec{c}] = 1 \]

### Final Answer

Thus, the value of the determinant \([ \vec{a} \, \vec{b} \, \vec{c} ]\) is:

\[ \boxed{1} \]