If `veca` is parallel to `vecb xx vecc, then (veca xx vecb) .(veca xx vecc) ` is equal to (a) `|veca|^(2)(vecb.vecc)` (b) `|vecb|^(2)(veca .vecc)` (c) `|vecc|^(2)(veca.vecb)` (d) none of these

Options

A. `|veca|^(2)(vecb.vecc)`
B. `|vecb|^(2)(veca .vecc)`
C. `|vecc|^(2)(veca.vecb)`
D. none of these

Explanation

To solve the problem, we need to analyze the given information and apply vector identities. Let's go through the steps systematically.

### Step 1: Understand the given condition
We are given that \(\vec{a}\) is parallel to \(\vec{b} \times \vec{c}\). This means we can express \(\vec{a}\) as:
\[ \vec{a} = \lambda (\vec{b} \times \vec{c}) \]
for some scalar \(\lambda\).

**Hint:** Recognize that parallel vectors can be expressed as scalar multiples of each other.

### Step 2: Set up the expression to evaluate
We need to evaluate \((\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c})\).

**Hint:** Recall the vector triple product identity: \(\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z}\).

### Step 3: Use the vector triple product identity
Let \(\vec{u} = \vec{a} \times \vec{c}\). Then, we can rewrite the expression:
\[ (\vec{a} \times \vec{b}) \cdot \vec{u} = \vec{a} \cdot (\vec{b} \times \vec{u}) \]
where \(\vec{u} = \vec{a} \times \vec{c}\).

**Hint:** This step simplifies the expression using the properties of the dot and cross products.

### Step 4: Expand using the triple product identity
Using the identity, we have:
\[ \vec{b} \times \vec{u} = \vec{b} \times (\vec{a} \times \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{c} \]

**Hint:** Substitute \(\vec{u}\) back into the expression to evaluate the dot product.

### Step 5: Substitute and simplify
Now, substituting back, we get:
\[ \vec{a} \cdot \left((\vec{b} \cdot \vec{c}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{c}\right) \]
This simplifies to:
\[ (\vec{b} \cdot \vec{c}) (\vec{a} \cdot \vec{a}) - (\vec{b} \cdot \vec{a})(\vec{a} \cdot \vec{c}) \]

**Hint:** Remember that \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\).

### Step 6: Use the fact that \(\vec{a}\) is parallel to \(\vec{b} \times \vec{c}\)
Since \(\vec{a}\) is parallel to \(\vec{b} \times \vec{c}\), we have \(\vec{a} \cdot \vec{b} = 0\). Therefore, the second term becomes zero:
\[ (\vec{b} \cdot \vec{c}) |\vec{a}|^2 \]

**Hint:** Recognize that the dot product being zero simplifies the expression significantly.

### Step 7: Final result
Thus, we conclude that:
\[ (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) = |\vec{a}|^2 (\vec{b} \cdot \vec{c}) \]

This matches option (a):
\[ |\vec{a}|^2 (\vec{b} \cdot \vec{c}) \]

### Conclusion
The correct answer is:
**(a) \(|\vec{a}|^2 (\vec{b} \cdot \vec{c})\)**