If a,b and c are three non-zero vectors such that no two of these are collinear. If the vector a+2b is collinear with c and b+3c is collinear with a(`lamda` being some non-zero scalar), then a+2b+6c is equal to

Options

A. A. 0
B. B. `lamdab`
C. C. `lamdac`
D. D. `lamda a`

Explanation

To solve the problem, we need to analyze the given conditions and derive the required expression step by step.

### Step 1: Set up the equations based on collinearity
We know that:
1. \( \mathbf{a} + 2\mathbf{b} \) is collinear with \( \mathbf{c} \).
This can be expressed as:
\[ \mathbf{a} + 2\mathbf{b} = \lambda \mathbf{c} \quad \text{(1)} \]
where \( \lambda \) is some non-zero scalar.

2. \( \mathbf{b} + 3\mathbf{c} \) is collinear with \( \mathbf{a} \).
This can be expressed as:
\[ \mathbf{b} + 3\mathbf{c} = \mu \mathbf{a} \quad \text{(2)} \]
where \( \mu \) is another non-zero scalar.

### Step 2: Express \( \mathbf{c} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \)
From equation (2), we can rearrange it to express \( \mathbf{c} \):
\[ 3\mathbf{c} = \mu \mathbf{a} - \mathbf{b} \]
\[ \mathbf{c} = \frac{\mu}{3} \mathbf{a} - \frac{1}{3} \mathbf{b} \quad \text{(3)} \]

### Step 3: Substitute \( \mathbf{c} \) into equation (1)
Now, we substitute equation (3) into equation (1):
\[ \mathbf{a} + 2\mathbf{b} = \lambda \left( \frac{\mu}{3} \mathbf{a} - \frac{1}{3} \mathbf{b} \right) \]
Expanding the right-hand side:
\[ \mathbf{a} + 2\mathbf{b} = \frac{\lambda \mu}{3} \mathbf{a} - \frac{\lambda}{3} \mathbf{b} \]

### Step 4: Rearranging the equation
Now, we can rearrange this equation:
\[ \mathbf{a} - \frac{\lambda \mu}{3} \mathbf{a} + 2\mathbf{b} + \frac{\lambda}{3} \mathbf{b} = 0 \]
Factoring out \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \left(1 - \frac{\lambda \mu}{3}\right) \mathbf{a} + \left(2 + \frac{\lambda}{3}\right) \mathbf{b} = 0 \]

### Step 5: Set coefficients to zero
Since \( \mathbf{a} \) and \( \mathbf{b} \) are non-collinear, the coefficients must be zero:
1. \( 1 - \frac{\lambda \mu}{3} = 0 \)
2. \( 2 + \frac{\lambda}{3} = 0 \)

### Step 6: Solve for \( \lambda \) and \( \mu \)
From the second equation:
\[ \frac{\lambda}{3} = -2 \implies \lambda = -6 \]

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

### Step 7: Find \( \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} \)
Now we substitute \( \lambda \) back into equation (1):
\[ \mathbf{a} + 2\mathbf{b} = -6\mathbf{c} \]
Adding \( 6\mathbf{c} \) to both sides:
\[ \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} = 0 \]

### Final Answer
Thus, we conclude:
\[ \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} = \mathbf{0} \]