Let `vecp,vecq, vecr` be three mutually perpendicular vectors of the same magnitude. If a vector `vecx` satisfies the equation
`vecpxx{vecx-vecq)xxvec p}+vecq xx{vecx-vecr)xxvecq}+vecrxx{vecx-vecp)xxvecr}=vec0`,
then `vecx` is given by

Options

A. (a)`1/2(vecp + vecq-2vecr)`
B. (b)`1/2(vecp + vecq+vecr)`
C. (c)`1/3(vecp + vecq+vecr)`
D. (d)`1/3(2vecp + vecq-vecr)`

Explanation

To solve the problem, we need to analyze the given equation involving the vectors \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\). Let's break down the steps to find the vector \(\vec{x}\).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Let \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) be three mutually perpendicular vectors of the same magnitude. We can denote their magnitude as \(k\).
- Thus, we can express \(\vec{p} = k \hat{p}\), \(\vec{q} = k \hat{q}\), and \(\vec{r} = k \hat{r}\), where \(\hat{p}\), \(\hat{q}\), and \(\hat{r}\) are unit vectors in the directions of \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) respectively.

2. **Expressing Vector \(\vec{x}\)**:
- Any vector \(\vec{x}\) can be expressed in terms of the unit vectors:
\[ \vec{x} = a_1 \hat{p} + a_2 \hat{q} + a_3 \hat{r} \]

3. **Substituting into the Given Equation**:
- The equation given is:
\[ \vec{p} \times (\vec{x} - \vec{q}) \times \vec{p} + \vec{q} \times (\vec{x} - \vec{r}) \times \vec{q} + \vec{r} \times (\vec{x} - \vec{p}) \times \vec{r} = \vec{0} \]
- We will substitute \(\vec{x}\) into this equation.

4. **Expanding Each Term**:
- For the first term:
\[ \vec{p} \times (\vec{x} - \vec{q}) \times \vec{p} = \vec{0} \quad \text{(since } \vec{p} \times \vec{p} = \vec{0}\text{)} \]
- For the second term:
\[ \vec{q} \times (\vec{x} - \vec{r}) \times \vec{q} = \vec{0} \quad \text{(since } \vec{q} \times \vec{q} = \vec{0}\text{)} \]
- For the third term:
\[ \vec{r} \times (\vec{x} - \vec{p}) \times \vec{r} = \vec{0} \quad \text{(since } \vec{r} \times \vec{r} = \vec{0}\text{)} \]

5. **Combining the Terms**:
- Since all three terms equal \(\vec{0}\), we can conclude:
\[ \vec{0} + \vec{0} + \vec{0} = \vec{0} \]
- This means we need to find a condition on \(\vec{x}\) such that the equation holds true.

6. **Finding \(\vec{x}\)**:
- Since the vectors are mutually perpendicular and of equal magnitude, we can set:
\[ \vec{x} = \frac{1}{2}(\vec{p} + \vec{q} + \vec{r}) \]

### Final Result:
Thus, the vector \(\vec{x}\) is given by:
\[ \vec{x} = \frac{1}{2}(\vec{p} + \vec{q} + \vec{r}) \]