`veca, vecb,vecc` are non-coplanar vectors and `vecp,vecq,vecr` are defined as `vecp = (vecb xx vecc)/([vecb vecc veca]),q=(vecc xx veca)/([vecc veca vecb]), vecr =(veca xx vecb)/([veca vecb vecc])` then `(veca + vecb).vecp+(vecb+vecc).vecq + (vecc + veca).vecr` is equal to.

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to evaluate the expression \((\vec{a} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r}\), where the vectors \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) are defined as follows:

\[ \vec{p} = \frac{\vec{b} \times \vec{c}}{[\vec{b}, \vec{c}, \vec{a}]}, \quad \vec{q} = \frac{\vec{c} \times \vec{a}}{[\vec{c}, \vec{a}, \vec{b}]}, \quad \vec{r} = \frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]} \]

### Step-by-step Solution:

1. **Expand the Expression**:
We start with the expression:
\[ (\vec{a} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r} \]
This can be expanded as:
\[ \vec{a} \cdot \vec{p} + \vec{b} \cdot \vec{p} + \vec{b} \cdot \vec{q} + \vec{c} \cdot \vec{q} + \vec{c} \cdot \vec{r} + \vec{a} \cdot \vec{r} \]

2. **Substituting the Values of \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\)**:
Substitute the definitions of \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\):
\[ \vec{a} \cdot \left(\frac{\vec{b} \times \vec{c}}{[\vec{b}, \vec{c}, \vec{a}]}\right) + \vec{b} \cdot \left(\frac{\vec{b} \times \vec{c}}{[\vec{b}, \vec{c}, \vec{a}]}\right) + \vec{b} \cdot \left(\frac{\vec{c} \times \vec{a}}{[\vec{c}, \vec{a}, \vec{b}]}\right) + \vec{c} \cdot \left(\frac{\vec{c} \times \vec{a}}{[\vec{c}, \vec{a}, \vec{b}]}\right) + \vec{c} \cdot \left(\frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]}\right) + \vec{a} \cdot \left(\frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]}\right) \]

3. **Using the Property of Dot and Cross Products**:
Recall that for any vector \(\vec{x}\), \(\vec{x} \cdot (\vec{y} \times \vec{z}) = 0\) if \(\vec{x}\) is one of the vectors in the cross product. Thus:
- \(\vec{b} \cdot (\vec{b} \times \vec{c}) = 0\)
- \(\vec{c} \cdot (\vec{c} \times \vec{a}) = 0\)
- \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\)

Therefore, we can simplify:
\[ \vec{a} \cdot \vec{p} + 0 + \vec{b} \cdot \vec{q} + 0 + \vec{c} \cdot \vec{r} + 0 \]

4. **Calculating \(\vec{a} \cdot \vec{p}\), \(\vec{b} \cdot \vec{q}\), and \(\vec{c} \cdot \vec{r}\)**:
Now we calculate:
\[ \vec{a} \cdot \vec{p} = \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{[\vec{b}, \vec{c}, \vec{a}]} = \frac{[\vec{a}, \vec{b}, \vec{c}]}{[\vec{b}, \vec{c}, \vec{a}]} \]
\[ \vec{b} \cdot \vec{q} = \frac{[\vec{b}, \vec{c}, \vec{a}]}{[\vec{c}, \vec{a}, \vec{b}]} \]
\[ \vec{c} \cdot \vec{r} = \frac{[\vec{c}, \vec{a}, \vec{b}]}{[\vec{a}, \vec{b}, \vec{c}]} \]

5. **Adding the Results**:
Since the determinants are cyclic, we have:
\[ \frac{[\vec{a}, \vec{b}, \vec{c}]}{[\vec{b}, \vec{c}, \vec{a}]} + \frac{[\vec{b}, \vec{c}, \vec{a}]}{[\vec{c}, \vec{a}, \vec{b}]} + \frac{[\vec{c}, \vec{a}, \vec{b}]}{[\vec{a}, \vec{b}, \vec{c}]} = 1 + 1 + 1 = 3 \]

### Final Result:
Thus, the value of the expression is:
\[ \boxed{3} \]