Let `veca , vecb, vec c ` be three non coplanar vectors , and let `vecp , vecq " and " vec r` be the vectors defined by the relation
`vecp = (vecb xx vec c )/([veca vecb vec c ]), vec q = (vec c xx vec a)/([veca vecb vec c ]) " and " vec r = (vec a xx vec b)/([veca vecb vec c ])`
Then the value of the expension
`(vec a + vec b) .vec p + (vecb + vec c) .q + (vec c + vec a) . vec r ` is equal to

Options

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

Answer

To solve the given 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:

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

### Step 1: Substitute the values of \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\)

Substituting these definitions into the expression gives:

\[ (\vec{a} + \vec{b}) \cdot \left(\frac{\vec{b} \times \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]}\right) + (\vec{b} + \vec{c}) \cdot \left(\frac{\vec{c} \times \vec{a}}{[\vec{a}, \vec{b}, \vec{c}]}\right) + (\vec{c} + \vec{a}) \cdot \left(\frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]}\right) \]

### Step 2: Factor out the common denominator

We can factor out the common denominator \([\vec{a}, \vec{b}, \vec{c}]\):

\[ \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} \left( (\vec{a} + \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} + \vec{a}) \cdot (\vec{a} \times \vec{b}) \right) \]

### Step 3: Evaluate each dot product

1. **First term**:
\[ (\vec{a} + \vec{b}) \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{c}) \]
The second term is zero because \(\vec{b} \cdot (\vec{b} \times \vec{c}) = 0\) (a vector is perpendicular to its cross product). Thus:
\[ (\vec{a} + \vec{b}) \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

2. **Second term**:
\[ (\vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a}) = \vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{c} \times \vec{a}) \]
The second term is zero for the same reason. Thus:
\[ (\vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a}) = \vec{b} \cdot (\vec{c} \times \vec{a}) \]

3. **Third term**:
\[ (\vec{c} + \vec{a}) \cdot (\vec{a} \times \vec{b}) = \vec{c} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{a} \times \vec{b}) \]
Again, the second term is zero. Thus:
\[ (\vec{c} + \vec{a}) \cdot (\vec{a} \times \vec{b}) = \vec{c} \cdot (\vec{a} \times \vec{b}) \]

### Step 4: Combine the results

Now we can combine all the results:

\[ \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} \left( \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{a} \times \vec{b}) \right) \]

### Step 5: Apply the scalar triple product identity

Using the identity of the scalar triple product, we know:

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

Thus, we have:

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{a} \times \vec{b}) = [\vec{a}, \vec{b}, \vec{c}] + [\vec{b}, \vec{c}, \vec{a}] + [\vec{c}, \vec{a}, \vec{b}] \]

Since all these determinants are equal:

\[ = 3[\vec{a}, \vec{b}, \vec{c}] \]

### Step 6: Final result

Thus, the expression simplifies to:

\[ \frac{3[\vec{a}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]} = 3 \]

### Conclusion

The final value of the expression is:

\[ \boxed{3} \]