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. `lamda a`
B. `lamdab`
C. `lamda c`
D. `0`

Explanation

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

### Step 1: Establish the first condition
We know that \( \mathbf{a} + 2\mathbf{b} \) is collinear with \( \mathbf{c} \). This means that there exists a scalar \( \lambda_1 \) such that:
\[ \mathbf{a} + 2\mathbf{b} = \lambda_1 \mathbf{c} \]
This can be rearranged to express \( \mathbf{c} \):
\[ \mathbf{c} = \frac{1}{\lambda_1} (\mathbf{a} + 2\mathbf{b}) \tag{1} \]

### Step 2: Establish the second condition
The second condition states that \( \mathbf{b} + 3\mathbf{c} \) is collinear with \( \mathbf{a} \). Thus, there exists a scalar \( \lambda_2 \) such that:
\[ \mathbf{b} + 3\mathbf{c} = \lambda_2 \mathbf{a} \]
Rearranging gives us:
\[ \mathbf{c} = \frac{1}{3} (\lambda_2 \mathbf{a} - \mathbf{b}) \tag{2} \]

### Step 3: Substitute equation (1) into equation (2)
Now we can substitute the expression for \( \mathbf{c} \) from equation (1) into equation (2):
\[ \frac{1}{\lambda_1} (\mathbf{a} + 2\mathbf{b}) = \frac{1}{3} (\lambda_2 \mathbf{a} - \mathbf{b}) \]

### Step 4: Clear the fractions
Multiplying through by \( 3\lambda_1 \) to eliminate the fractions gives:
\[ 3(\mathbf{a} + 2\mathbf{b}) = \lambda_2 \mathbf{a} - \mathbf{b} \]

### Step 5: Rearranging the equation
Rearranging the equation results in:
\[ 3\mathbf{a} + 6\mathbf{b} + \mathbf{b} - \lambda_2 \mathbf{a} = 0 \]
This simplifies to:
\[ (3 - \lambda_2) \mathbf{a} + 7\mathbf{b} = 0 \]

### Step 6: Coefficient comparison
Since \( \mathbf{a} \) and \( \mathbf{b} \) are not collinear, the coefficients must independently equal zero:
1. \( 3 - \lambda_2 = 0 \) implies \( \lambda_2 = 3 \)
2. \( 7 = 0 \) gives no new information.

### Step 7: Substitute \( \lambda_2 \) back
Now we can substitute \( \lambda_2 = 3 \) back into equation (2):
\[ \mathbf{c} = \frac{1}{3} (3\mathbf{a} - \mathbf{b}) = \mathbf{a} - \frac{1}{3}\mathbf{b} \]

### Step 8: Find \( \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} \)
Now we can find \( \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} \):
\[ \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} = \mathbf{a} + 2\mathbf{b} + 6\left(\mathbf{a} - \frac{1}{3}\mathbf{b}\right) \]
This simplifies to:
\[ \mathbf{a} + 2\mathbf{b} + 6\mathbf{a} - 2\mathbf{b} = 7\mathbf{a} \]

### Final Result
Thus, the final result is:
\[ \mathbf{a} + 2\mathbf{b} + 6\mathbf{c} = 7\mathbf{a} \]