Let a,b and c be three non-zero vectors which are pairwise non-collinear. If a+3b is collinear with c and b+2c is collinear with a, then a+3b+6c is

Options

A. a+c
B. a
C. `c`
D. 0

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a clearer structure.

### Step 1: Understanding Collinearity
Given that \( a + 3b \) is collinear with \( c \), we can express this relationship mathematically as:
\[ a + 3b = \lambda c \]
for some scalar \( \lambda \).

### Step 2: Another Collinearity Condition
Similarly, since \( b + 2c \) is collinear with \( a \), we can express this as:
\[ b + 2c = \mu a \]
for some scalar \( \mu \).

### Step 3: Rearranging the First Equation
From the first equation, we can rearrange it to isolate \( c \):
\[ c = \frac{1}{\lambda}(a + 3b) \]

### Step 4: Rearranging the Second Equation
From the second equation, we can isolate \( a \):
\[ a = \frac{1}{\mu}(b + 2c) \]

### Step 5: Substituting for \( c \) in the Second Equation
Substituting the expression for \( c \) from Step 3 into the equation for \( a \):
\[ a = \frac{1}{\mu}\left(b + 2\left(\frac{1}{\lambda}(a + 3b)\right)\right) \]
This simplifies to:
\[ a = \frac{1}{\mu}\left(b + \frac{2}{\lambda}(a + 3b)\right) \]
Expanding this gives:
\[ a = \frac{1}{\mu}\left(b + \frac{2}{\lambda}a + \frac{6}{\lambda}b\right) \]
Combining like terms:
\[ a = \frac{1}{\mu}\left(\left(1 + \frac{6}{\lambda}\right)b + \frac{2}{\lambda}a\right) \]

### Step 6: Isolating \( a \)
Rearranging gives:
\[ a - \frac{2}{\mu\lambda}a = \frac{1 + \frac{6}{\lambda}}{\mu}b \]
Factoring out \( a \):
\[ a\left(1 - \frac{2}{\mu\lambda}\right) = \frac{1 + \frac{6}{\lambda}}{\mu}b \]

### Step 7: Finding \( a + 3b + 6c \)
Now, we need to find \( a + 3b + 6c \):
\[ a + 3b + 6c = a + 3b + 6\left(\frac{1}{\lambda}(a + 3b)\right) \]
This simplifies to:
\[ = a + 3b + \frac{6}{\lambda}a + \frac{18}{\lambda}b \]
Combining terms gives:
\[ \left(1 + \frac{6}{\lambda}\right)a + \left(3 + \frac{18}{\lambda}\right)b \]

### Step 8: Setting the Result Equal to Zero
Since \( a, b, c \) are pairwise non-collinear, the coefficients of \( a \) and \( b \) must equal zero:
\[ 1 + \frac{6}{\lambda} = 0 \quad \text{and} \quad 3 + \frac{18}{\lambda} = 0 \]

### Step 9: Solving for \( \lambda \)
From the first equation:
\[ \frac{6}{\lambda} = -1 \implies \lambda = -6 \]
From the second equation:
\[ \frac{18}{\lambda} = -3 \implies \lambda = -6 \]

### Conclusion
Thus, substituting \( \lambda = -6 \) back into our expression for \( a + 3b + 6c \):
\[ a + 3b + 6c = 0 \]

### Final Answer
\[ \boxed{0} \]