If `vecalpha||(vecbxxvecgamma), then (vecalphaxxvecbeta).(vecalphaxxvecgamma)=` (A) `|vecalpha|^2(vecbeta.vecgamma)` (B) `|vecbeta|^2(vecgamma.vecalpha)` (C) `|vecgamma|^2(vecalpha.vecbeta)` (D) `|vecalpha||vecbeta||vecgamma|`

Options

A. `|vecalpha|^(2)(vecbeta.vecgamma)`
B. `|vecbeta|^(2)(vecgamma.vecalpha)`
C. `|vecgamma|^(2)(vecalpha.vecbeta)`
D. `|vecalpha||vecbeta||vecgamma|`

Explanation

To solve the problem, we start with the given condition that \(\vec{\alpha} \parallel (\vec{\beta} \times \vec{\gamma})\). This means that \(\vec{\alpha}\) is parallel to the vector resulting from the cross product of \(\vec{\beta}\) and \(\vec{\gamma}\).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
Since \(\vec{\alpha} \parallel (\vec{\beta} \times \vec{\gamma})\), we can express this as:
\[ \vec{\alpha} = k (\vec{\beta} \times \vec{\gamma}) \]
for some scalar \(k\).

2. **Cross Product**:
We need to find \((\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma})\).

3. **Using the Vector Triple Product Identity**:
We can use the vector triple product identity:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Applying this, we can rewrite \(\vec{\alpha} \times \vec{\beta}\) and \(\vec{\alpha} \times \vec{\gamma}\).

4. **Calculating the Dot Product**:
We can express the dot product as:
\[ (\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma}) = |\vec{\alpha}|^2 (\vec{\beta} \cdot \vec{\gamma}) - (\vec{\alpha} \cdot \vec{\beta})(\vec{\alpha} \cdot \vec{\gamma}) \]

5. **Using Perpendicularity**:
Since \(\vec{\alpha}\) is perpendicular to both \(\vec{\beta}\) and \(\vec{\gamma}\) (as derived from the condition), we have:
\[ \vec{\alpha} \cdot \vec{\beta} = 0 \quad \text{and} \quad \vec{\alpha} \cdot \vec{\gamma} = 0 \]
Therefore, the terms involving the dot products become zero:
\[ (\vec{\alpha} \cdot \vec{\beta})(\vec{\alpha} \cdot \vec{\gamma}) = 0 \]

6. **Final Expression**:
Thus, we simplify to:
\[ (\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma}) = |\vec{\alpha}|^2 (\vec{\beta} \cdot \vec{\gamma}) \]

7. **Conclusion**:
Therefore, the final answer is:
\[ (\vec{\alpha} \times \vec{\beta}) \cdot (\vec{\alpha} \times \vec{\gamma}) = |\vec{\alpha}|^2 (\vec{\beta} \cdot \vec{\gamma}) \]
This matches option (A).