If `veca, vecb and vecc` are three non-zero vectors, no two of which are collinear, `veca +2 vecb` is collinear with `vecc and vecb + 3 vecc` is collinear with `veca`, then find the value of `|veca + 2vecb + 6vecc|`.

Options

A. `veca + vecc`
B. `veca`
C. `vecc`
D. `vec0`

Answer

To solve the problem, we need to analyze the given conditions regarding the vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \).

### Step 1: Set up the equations based on collinearity
From the problem, we have two conditions:

1. \( \vec{a} + 2\vec{b} \) is collinear with \( \vec{c} \).
2. \( \vec{b} + 3\vec{c} \) is collinear with \( \vec{a} \).

This means we can express these relationships mathematically as:
- \( \vec{a} + 2\vec{b} = \lambda \vec{c} \) for some scalar \( \lambda \).
- \( \vec{b} + 3\vec{c} = \mu \vec{a} \) for some scalar \( \mu \).

### Step 2: Express \( \vec{a} \) in terms of \( \vec{b} \) and \( \vec{c} \)
From the first equation, we can express \( \vec{a} \):
\[ \vec{a} = \lambda \vec{c} - 2\vec{b} \]

### Step 3: Substitute \( \vec{a} \) into the second equation
Now we substitute this expression for \( \vec{a} \) into the second equation:
\[ \vec{b} + 3\vec{c} = \mu (\lambda \vec{c} - 2\vec{b}) \]

### Step 4: Expand and rearrange the equation
Expanding the right-hand side gives:
\[ \vec{b} + 3\vec{c} = \mu \lambda \vec{c} - 2\mu \vec{b} \]

Now, rearranging this equation:
\[ \vec{b} + 2\mu \vec{b} = \mu \lambda \vec{c} - 3\vec{c} \]
\[ (1 + 2\mu) \vec{b} = (\mu \lambda - 3) \vec{c} \]

### Step 5: Compare coefficients
Since \( \vec{b} \) and \( \vec{c} \) are not collinear, we can equate the coefficients:
1. \( 1 + 2\mu = 0 \)
2. \( \mu \lambda - 3 = 0 \)

From the first equation:
\[ 2\mu = -1 \implies \mu = -\frac{1}{2} \]

Substituting \( \mu \) into the second equation:
\[ -\frac{1}{2} \lambda - 3 = 0 \implies -\frac{1}{2} \lambda = 3 \implies \lambda = -6 \]

### Step 6: Substitute back to find \( \vec{a} + 2\vec{b} + 6\vec{c} \)
Now we substitute \( \lambda \) back into the first equation:
\[ \vec{a} + 2\vec{b} = -6\vec{c} \]

Adding \( 6\vec{c} \) to both sides:
\[ \vec{a} + 2\vec{b} + 6\vec{c} = -6\vec{c} + 6\vec{c} = \vec{0} \]

### Final Result
Thus, we find:
\[ |\vec{a} + 2\vec{b} + 6\vec{c}| = |\vec{0}| = 0 \]