Let `veca,vecb,vecc` be three mutually perpendicular vectors having same magnitude and `vecr` is a vector satisfying
`vecaxx((vecr-vecb)xxveca)+vecbxx((vecr-vecc)xxvecb)+veccxx((vecr-veca)xxvecc)=vec0,` then `vecr` is equal to

Options

A. A.`1/3(veca+vecb+vecc)`
B. B. `1/2(veca+vecb+vecc)`
C. C. `3/2(veca+vecb+vecc)`
D. D.`2(veca+vecb+vecc)`

Explanation

To solve the problem, we need to analyze the given vector equation step by step.

### Given:
Let \(\vec{a}, \vec{b}, \vec{c}\) be three mutually perpendicular vectors having the same magnitude. We need to find the vector \(\vec{r}\) that satisfies the equation:

\[ \vec{a} \times ((\vec{r} - \vec{b}) \times \vec{a}) + \vec{b} \times ((\vec{r} - \vec{c}) \times \vec{b}) + \vec{c} \times ((\vec{r} - \vec{a}) \times \vec{c}) = \vec{0} \]

### Step 1: Understanding the Cross Product
Since \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular, we can use the properties of the cross product. The cross product of any vector with itself is zero, and the cross product of two perpendicular vectors gives a vector perpendicular to both.

### Step 2: Expand Each Term
We will expand each term in the equation using the distributive property of the cross product.

1. **First Term**:
\[ \vec{a} \times ((\vec{r} - \vec{b}) \times \vec{a}) = \vec{a} \times (\vec{r} \times \vec{a} - \vec{b} \times \vec{a}) = \vec{a} \times (\vec{r} \times \vec{a}) - \vec{b} \]
(Since \(\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}\))

2. **Second Term**:
\[ \vec{b} \times ((\vec{r} - \vec{c}) \times \vec{b}) = \vec{b} \times (\vec{r} \times \vec{b} - \vec{c} \times \vec{b}) = \vec{b} \times (\vec{r} \times \vec{b}) - \vec{c} \]

3. **Third Term**:
\[ \vec{c} \times ((\vec{r} - \vec{a}) \times \vec{c}) = \vec{c} \times (\vec{r} \times \vec{c} - \vec{a} \times \vec{c}) = \vec{c} \times (\vec{r} \times \vec{c}) - \vec{a} \]

### Step 3: Combine the Terms
Now, we combine all the expanded terms:
\[ \vec{a} \times (\vec{r} \times \vec{a}) + \vec{b} \times (\vec{r} \times \vec{b}) + \vec{c} \times (\vec{r} \times \vec{c}) - \vec{b} - \vec{c} - \vec{a} = \vec{0} \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ \vec{a} \times (\vec{r} \times \vec{a}) + \vec{b} \times (\vec{r} \times \vec{b}) + \vec{c} \times (\vec{r} \times \vec{c}) = \vec{a} + \vec{b} + \vec{c} \]

### Step 5: Using the Vector Identity
Using the vector triple product identity, we have:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Applying this to each term, we can simplify further.

### Step 6: Set \(\vec{r}\)
Assuming \(\vec{r} = k \vec{a} + k \vec{b} + k \vec{c}\) for some scalar \(k\), we can substitute this back into our equation and solve for \(k\).

### Step 7: Conclusion
After evaluating, we find that \(k = 0\) leads to \(\vec{r} = \vec{0}\).

Thus, the solution is:
\[ \vec{r} = \vec{0} \]