If `veca , vecb and vecc` are three non-zero, non- coplanar vectors and `vecb_(1)=vecb-(vecb.veca)/(|veca|^(2))veca, \ vecb_(2)=vecb+(vecb.veca)/(|veca|^(2))veca, \ vecc_(1)=vecc-(vecc.veca)/(|veca|^(2))veca+ (vecb.vecc)/(|vecc|^(2))vecb_(1),` `vecc_(2)=vecc-(vecc.veca)/(|veca|^(2)) veca-(vecbvecc)/(|vecb_(1)|^(2))vecb_(1), \ vecc_(3)=vecc- (vecc.veca)/(|vecc|^(2))veca + (vecb.vecc)/(|vecc|^(2))vecb_(1),` ` vecc_(4)=vecc - (vecc.veca)/(|vecc|^(2))veca= (vecb.vecc)/(|vecb|^(2))vecb_(1)`, then the set of mutually orthogonal vectors is

Options

A. (a) `(veca,vecb_(1),vecc_(3))`
B. (b) `(veca,vecb_(1),vecc_(2))`
C. (c) `(veca, vecb_(1),vecc_(1))`
D. (d) `(veca,vecb_(2),vecc_(2))`

Explanation

To solve the problem, we need to analyze the given vectors and check for mutual orthogonality among them. We have three non-zero, non-coplanar vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \), and we need to find the set of mutually orthogonal vectors among the defined vectors \( \vec{b}_1, \vec{b}_2, \vec{c}_1, \vec{c}_2, \vec{c}_3, \vec{c}_4 \).

### Step-by-Step Solution:

1. **Define the Vectors**:
- Given:
\[ \vec{b}_1 = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} \]
\[ \vec{b}_2 = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} \]
\[ \vec{c}_1 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} + \frac{\vec{b} \cdot \vec{c}}{|\vec{c}|^2} \vec{b}_1 \]
\[ \vec{c}_2 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} \vec{b}_1 \]
\[ \vec{c}_3 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2} \vec{a} + \frac{\vec{b} \cdot \vec{c}}{|\vec{c}|^2} \vec{b}_1 \]
\[ \vec{c}_4 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2} \vec{a} = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2} \vec{b}_1 \]

2. **Check Orthogonality**:
- We need to check if the dot products among these vectors equal zero.
- Start with \( \vec{a} \cdot \vec{b}_1 \):
\[ \vec{a} \cdot \vec{b}_1 = \vec{a} \cdot \left( \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} \right) \]
\[ = \vec{a} \cdot \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} (\vec{a} \cdot \vec{a}) = \vec{a} \cdot \vec{b} - \frac{(\vec{b} \cdot \vec{a}) |\vec{a}|^2}{|\vec{a}|^2} = 0 \]
- Thus, \( \vec{a} \) and \( \vec{b}_1 \) are orthogonal.

3. **Check \( \vec{b}_1 \) with \( \vec{c}_2 \)**:
- Calculate \( \vec{b}_1 \cdot \vec{c}_2 \):
\[ \vec{b}_1 \cdot \vec{c}_2 = \vec{b}_1 \cdot \left( \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} \vec{b}_1 \right) \]
- Simplifying gives:
\[ = \vec{b}_1 \cdot \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{b}_1 \cdot \vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} |\vec{b}_1|^2 \]
- Since \( \vec{b}_1 \cdot \vec{a} = 0 \) and \( \vec{b}_1 \cdot \vec{b}_1 = |\vec{b}_1|^2 \), we find:
\[ \vec{b}_1 \cdot \vec{c}_2 = 0 \]

4. **Check \( \vec{a} \cdot \vec{c}_2 \)**:
- Calculate \( \vec{a} \cdot \vec{c}_2 \):
\[ \vec{a} \cdot \vec{c}_2 = \vec{a} \cdot \left( \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} \vec{b}_1 \right) \]
- Simplifying gives:
\[ = \vec{a} \cdot \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} |\vec{a}|^2 - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} \cdot 0 = 0 \]

5. **Conclusion**:
- We have shown that:
\[ \vec{a} \cdot \vec{b}_1 = 0, \quad \vec{b}_1 \cdot \vec{c}_2 = 0, \quad \vec{a} \cdot \vec{c}_2 = 0 \]
- Therefore, the set of mutually orthogonal vectors is \( \{ \vec{a}, \vec{b}_1, \vec{c}_2 \} \).

### Final Answer:
The set of mutually orthogonal vectors is \( \{ \vec{a}, \vec{b}_1, \vec{c}_2 \} \).