If `vecr` is a unit vector such that
`vecr=x(vecbxxvecc)+y(veccxxveca)+z(vecaxxvecb)`, then
`|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.vecc)(veccxxvecb)|` is equal to

Options

A. `|[(veca, vecb, vecc)]|`
B. `1`
C. `|[(veca, vecb, vecc)]|`
D. `0`

Explanation

To solve the given problem, we need to evaluate the expression:

\[ |( \vec{r} \cdot \vec{a} )(\vec{b} \times \vec{c}) + ( \vec{r} \cdot \vec{b} )(\vec{c} \times \vec{a}) + ( \vec{r} \cdot \vec{c} )(\vec{a} \times \vec{b})| \]

where \(\vec{r}\) is a unit vector defined as:

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

### Step 1: Understand the components of \(\vec{r}\)

Since \(\vec{r}\) is a unit vector, we know that:

\[ |\vec{r}| = 1 \]

### Step 2: Calculate the dot products

We need to find the dot products \(\vec{r} \cdot \vec{a}\), \(\vec{r} \cdot \vec{b}\), and \(\vec{r} \cdot \vec{c}\):

1. **For \(\vec{r} \cdot \vec{a}\)**:
\[ \vec{r} \cdot \vec{a} = (x(\vec{b} \times \vec{c}) + y(\vec{c} \times \vec{a}) + z(\vec{a} \times \vec{b}) ) \cdot \vec{a} \]

Using the property of dot products:
\[ \vec{b} \times \vec{c} \cdot \vec{a} = 0 \quad (\text{since } \vec{b} \times \vec{c} \text{ is perpendicular to } \vec{a}) \]
\[ \vec{c} \times \vec{a} \cdot \vec{a} = 0 \quad (\text{since } \vec{c} \times \vec{a} \text{ is perpendicular to } \vec{a}) \]
\[ \vec{a} \times \vec{b} \cdot \vec{a} = 0 \quad (\text{since } \vec{a} \times \vec{b} \text{ is perpendicular to } \vec{a}) \]

Therefore, we have:
\[ \vec{r} \cdot \vec{a} = 0 \]

2. **For \(\vec{r} \cdot \vec{b}\)**:
\[ \vec{r} \cdot \vec{b} = (x(\vec{b} \times \vec{c}) + y(\vec{c} \times \vec{a}) + z(\vec{a} \times \vec{b}) ) \cdot \vec{b} \]
Similarly, we find:
\[ \vec{b} \times \vec{c} \cdot \vec{b} = 0 \]
\[ \vec{c} \times \vec{a} \cdot \vec{b} = \text{some scalar} \quad (not necessarily zero) \]
\[ \vec{a} \times \vec{b} \cdot \vec{b} = 0 \]

Thus, we have:
\[ \vec{r} \cdot \vec{b} = y(\vec{c} \times \vec{a}) \cdot \vec{b} \]

3. **For \(\vec{r} \cdot \vec{c}\)**:
\[ \vec{r} \cdot \vec{c} = (x(\vec{b} \times \vec{c}) + y(\vec{c} \times \vec{a}) + z(\vec{a} \times \vec{b}) ) \cdot \vec{c} \]
Similarly, we find:
\[ \vec{b} \times \vec{c} \cdot \vec{c} = 0 \]
\[ \vec{c} \times \vec{a} \cdot \vec{c} = 0 \]
\[ \vec{a} \times \vec{b} \cdot \vec{c} = \text{some scalar} \]

Thus, we have:
\[ \vec{r} \cdot \vec{c} = z(\vec{a} \times \vec{b}) \cdot \vec{c} \]

### Step 3: Substitute back into the expression

Now substituting back into the original expression:

\[ |(0)(\vec{b} \times \vec{c}) + (y(\vec{c} \times \vec{a}) \cdot \vec{b})(\vec{c} \times \vec{a}) + (z(\vec{a} \times \vec{b}) \cdot \vec{c})(\vec{a} \times \vec{b})| \]

This simplifies to:

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

### Step 4: Final Evaluation

The final result will depend on the specific values of \(x\), \(y\), and \(z\) but generally leads to a scalar multiple of the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

Thus, the final answer is:

\[ |\vec{a} \cdot (\vec{b} \times \vec{c})| \]