Let `veca,vecb , vecc` are three vectors of which every pair is non collinear , and the vectors `veca+3vecb` and `2vecb+vecc` are collinear with `vecc` are `veca` respectively. If `vecb.vecb=1`, then find `|2veca+3vecc|`.

Answer

To solve the problem, we need to find the magnitude of the vector \( |2\vec{a} + 3\vec{c}| \) given the conditions about the vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \).

### Step-by-step Solution:

1. **Understanding the Given Conditions**:
- We know that \( \vec{a} + 3\vec{b} \) is collinear with \( \vec{c} \).
- We also know that \( 2\vec{b} + \vec{c} \) is collinear with \( \vec{a} \).
- Additionally, we are given that \( \vec{b} \cdot \vec{b} = 1 \).

2. **Expressing Collinearity**:
- Since \( \vec{a} + 3\vec{b} \) is collinear with \( \vec{c} \), we can write:
\[ \vec{a} + 3\vec{b} = \lambda \vec{c} \quad (1) \]
for some scalar \( \lambda \).
- Similarly, since \( 2\vec{b} + \vec{c} \) is collinear with \( \vec{a} \), we can write:
\[ 2\vec{b} + \vec{c} = \mu \vec{a} \quad (2) \]
for some scalar \( \mu \).

3. **Substituting and Rearranging**:
- From equation (1), we can express \( \vec{c} \):
\[ \vec{c} = \frac{1}{\lambda} (\vec{a} + 3\vec{b}) \]
- From equation (2), we can express \( \vec{c} \) as well:
\[ \vec{c} = \mu \vec{a} - 2\vec{b} \]

4. **Equating the Two Expressions for \( \vec{c} \)**:
- Setting the two expressions for \( \vec{c} \) equal gives:
\[ \frac{1}{\lambda} (\vec{a} + 3\vec{b}) = \mu \vec{a} - 2\vec{b} \]
- Rearranging this, we get:
\[ \vec{a} + 3\vec{b} = \lambda (\mu \vec{a} - 2\vec{b}) \]

5. **Expanding and Collecting Like Terms**:
- Expanding and rearranging leads to:
\[ \vec{a} + 3\vec{b} = \lambda \mu \vec{a} - 2\lambda \vec{b} \]
- Rearranging gives:
\[ (1 - \lambda \mu) \vec{a} + (3 + 2\lambda) \vec{b} = 0 \]

6. **Using Non-Collinearity**:
- Since \( \vec{a} \) and \( \vec{b} \) are non-collinear, we can set the coefficients to zero:
\[ 1 - \lambda \mu = 0 \quad \text{and} \quad 3 + 2\lambda = 0 \]

7. **Solving for \( \lambda \) and \( \mu \)**:
- From \( 3 + 2\lambda = 0 \), we find:
\[ \lambda = -\frac{3}{2} \]
- Substituting \( \lambda \) into \( 1 - \lambda \mu = 0 \):
\[ 1 + \frac{3}{2} \mu = 0 \implies \mu = -\frac{2}{3} \]

8. **Finding \( 2\vec{a} + 3\vec{c} \)**:
- Substitute \( \lambda \) back into either expression for \( \vec{c} \):
\[ \vec{c} = -\frac{3}{2} \vec{c} + 3\vec{b} \]
- Thus, we can find \( 2\vec{a} + 3\vec{c} \):
\[ 2\vec{a} + 3\vec{c} = 6\vec{b} \]

9. **Calculating the Magnitude**:
- The magnitude is given by:
\[ |2\vec{a} + 3\vec{c}| = |6\vec{b}| \]
- Since \( \vec{b} \cdot \vec{b} = 1 \), we have \( |\vec{b}| = 1 \).
- Therefore:
\[ |6\vec{b}| = 6 \cdot |\vec{b}| = 6 \cdot 1 = 6 \]

### Final Answer:
\[ |2\vec{a} + 3\vec{c}| = 6 \]