If `[veca vecb vecc]=1` then value of `(veca.vecbxxvecc)/(veccxxveca.vecb)+(vecb.veccxxveca)/(vecaxxvecb.vecc)+(vecc.vecaxxvecb)/(vecbxxvecc.veca)` is

Options

A. `3`
B. `1`
C. `-1`
D. None of these

Explanation

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

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

Given that \([\vec{a} \, \vec{b} \, \vec{c}] = 1\), we know that the scalar triple product of the vectors \(\vec{a}, \vec{b}, \vec{c}\) is equal to 1.

### Step 1: Rewrite the Scalar Triple Products
The scalar triple product can be expressed in terms of determinants. Therefore, we can write:

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a} \, \vec{b} \, \vec{c}] = 1 \]

Similarly, we can express the other terms:

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

\[ \vec{b} \cdot (\vec{c} \times \vec{a}) = [\vec{b} \, \vec{c} \, \vec{a}] = 1 \]

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

\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = [\vec{c} \, \vec{a} \, \vec{b}] = 1 \]

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

### Step 2: Substitute the Values
Now substituting these values into the expression:

\[ \frac{1}{1} + \frac{1}{1} + \frac{1}{1} \]

### Step 3: Simplify the Expression
This simplifies to:

\[ 1 + 1 + 1 = 3 \]

### Conclusion
Thus, the value of the given expression is:

\[ \boxed{3} \]