If `a(vecalphaxxvecbeta)=b(vecbetaxxvecgamma)+c(vecgammaxxvecalpha)=vec0` and at least one of a,b and c is non zero then vectors `vecalpha, vecbeta, vecgamma` are (A) parallel (B) coplanar (C) mutually perpendicular (D) none of these

Options

A. parallel
B. coplanar
C. mutually perpendicular
D. none of these

Explanation

To solve the problem, we start with the given equation:

\[ a(\vec{\alpha} \times \vec{\beta}) = b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \vec{0} \]

We need to analyze this equation to determine the relationship between the vectors \(\vec{\alpha}\), \(\vec{\beta}\), and \(\vec{\gamma}\).

### Step 1: Understanding the Equation
The equation states that a linear combination of the cross products of the vectors equals the zero vector. This indicates that the vectors involved may be related in such a way that they lie in the same plane.

### Step 2: Taking the Dot Product with \(\vec{\gamma}\)
We take the dot product of the entire equation with \(\vec{\gamma}\):

\[ a(\vec{\alpha} \times \vec{\beta}) \cdot \vec{\gamma} + b(\vec{\beta} \times \vec{\gamma}) \cdot \vec{\gamma} + c(\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\gamma} = 0 \]

The term \(b(\vec{\beta} \times \vec{\gamma}) \cdot \vec{\gamma}\) becomes zero because the cross product of any vector with itself is zero. Thus, we simplify to:

\[ a(\vec{\alpha} \times \vec{\beta}) \cdot \vec{\gamma} + c(\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\gamma} = 0 \]

### Step 3: Further Simplifying
The term \(c(\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\gamma}\) also becomes zero for the same reason. Therefore, we have:

\[ a(\vec{\alpha} \times \vec{\beta}) \cdot \vec{\gamma} = 0 \]

This implies that either \(a = 0\) or \((\vec{\alpha} \times \vec{\beta}) \cdot \vec{\gamma} = 0\).

### Step 4: Taking the Dot Product with \(\vec{\beta}\)
Next, we take the dot product of the original equation with \(\vec{\beta}\):

\[ a(\vec{\alpha} \times \vec{\beta}) \cdot \vec{\beta} + b(\vec{\beta} \times \vec{\gamma}) \cdot \vec{\beta} + c(\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\beta} = 0 \]

Again, the term \(b(\vec{\beta} \times \vec{\gamma}) \cdot \vec{\beta}\) becomes zero. Thus, we simplify to:

\[ c(\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\beta} = 0 \]

This implies that either \(c = 0\) or \((\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\beta} = 0\).

### Step 5: Taking the Dot Product with \(\vec{\alpha}\)
Finally, we take the dot product of the original equation with \(\vec{\alpha}\):

\[ a(\vec{\alpha} \times \vec{\beta}) \cdot \vec{\alpha} + b(\vec{\beta} \times \vec{\gamma}) \cdot \vec{\alpha} + c(\vec{\gamma} \times \vec{\alpha}) \cdot \vec{\alpha} = 0 \]

The term \(a(\vec{\alpha} \times \vec{\beta}) \cdot \vec{\alpha}\) becomes zero. Thus, we simplify to:

\[ b(\vec{\beta} \times \vec{\gamma}) \cdot \vec{\alpha} = 0 \]

This implies that either \(b = 0\) or \((\vec{\beta} \times \vec{\gamma}) \cdot \vec{\alpha} = 0\).

### Conclusion
Since at least one of \(a\), \(b\), or \(c\) is non-zero, and we have derived conditions that lead to the conclusion that the scalar triple product \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}] = 0\), we can conclude that the vectors \(\vec{\alpha}\), \(\vec{\beta}\), and \(\vec{\gamma}\) are coplanar.

Thus, the answer is:
**(B) coplanar.**