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 orthogonal vectors is

Options

A. `(veca,vecb_(1),vecc_(3))`
B. `(vecca,vecb_(1),vecc_(2))`
C. `(veca, vecb_(1),vecc_(1))`
D. `(veca,vecb_(2),vecc_(2))`

Explanation

To solve the problem, we need to analyze the given vectors and their relationships. We are tasked with determining which set of vectors is orthogonal. Let's break down the solution step by step.

### Step 1: Understanding the Definitions
We have three non-zero, non-coplanar vectors: \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). We also have derived vectors \(\vec{b}_1\), \(\vec{b}_2\), \(\vec{c}_1\), \(\vec{c}_2\), \(\vec{c}_3\), and \(\vec{c}_4\) based on the given formulas.

### Step 2: Calculate \(\vec{b}_1\)
\[ \vec{b}_1 = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} \]
This represents the projection of \(\vec{b}\) onto \(\vec{a}\) subtracted from \(\vec{b}\), which gives us a vector orthogonal to \(\vec{a}\).

### Step 3: Calculate \(\vec{c}_2\)
\[ \vec{c}_2 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} \vec{b}_1 \]
This vector \(\vec{c}_2\) is adjusted similarly to \(\vec{c}\) by subtracting projections onto \(\vec{a}\) and \(\vec{b}_1\).

### Step 4: Check Orthogonality
To check if the vectors are orthogonal, we need to compute the dot products:

1. **Check \(\vec{a} \cdot \vec{b}_1\)**:
\[ \vec{a} \cdot \vec{b}_1 = \vec{a} \cdot \left(\vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}\right) = \vec{a} \cdot \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} \cdot \vec{a} = \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} = 0 \]
Thus, \(\vec{a}\) is orthogonal to \(\vec{b}_1\).

2. **Check \(\vec{a} \cdot \vec{c}_2\)**:
\[ \vec{a} \cdot \vec{c}_2 = \vec{a} \cdot \left(\vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}_1|^2} \vec{b}_1\right) \]
The first term will simplify to zero as shown previously. The second term also simplifies to zero due to the orthogonality established earlier. Thus, \(\vec{a} \cdot \vec{c}_2 = 0\).

3. **Check \(\vec{b}_1 \cdot \vec{c}\)**:
\[ \vec{b}_1 \cdot \vec{c} = \left(\vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}\right) \cdot \vec{c} \]
The first term gives a non-zero value, but the second term will also simplify to zero since \(\vec{b}_1\) is orthogonal to \(\vec{a}\). Therefore, \(\vec{b}_1 \cdot \vec{c} = 0\).

### Conclusion
From the calculations above, we find that the vectors \(\vec{a}\), \(\vec{b}_1\), and \(\vec{c}_2\) are orthogonal to each other. Therefore, the set of orthogonal vectors is:

\[ \{\vec{a}, \vec{b}_1, \vec{c}_2\} \]