Let a, b, c be three vectors `(ne 0`), no two of which are collinear . If a + 2b is collinear with c, b + 3c is collinear with a then `a + 2b + 6c` is

Options

A. parallel to a
B. parallel to b
C. parallel to c
D. 0

Answer

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

### Step-by-Step Solution:

1. **Understanding Collinearity**:
- Two vectors \( u \) and \( v \) are said to be collinear if there exists a scalar \( k \) such that \( u = k v \).
- Given that \( a + 2b \) is collinear with \( c \), we can write:
\[ a + 2b = k_1 c \quad (1) \]
for some scalar \( k_1 \).

2. **Second Condition**:
- The second condition states that \( b + 3c \) is collinear with \( a \). Thus, we can write:
\[ b + 3c = k_2 a \quad (2) \]
for some scalar \( k_2 \).

3. **Substituting from (2) into (1)**:
- From equation (2), we can express \( b \) in terms of \( a \) and \( c \):
\[ b = k_2 a - 3c \quad (3) \]
- Substitute equation (3) into equation (1):
\[ a + 2(k_2 a - 3c) = k_1 c \]
Simplifying this gives:
\[ a + 2k_2 a - 6c = k_1 c \]
\[ (1 + 2k_2)a = (k_1 + 6)c \quad (4) \]

4. **Analyzing the Result**:
- From equation (4), we see that \( a \) and \( c \) cannot be collinear (as given in the problem). Therefore, the coefficients of \( a \) and \( c \) must both equal zero:
\[ 1 + 2k_2 = 0 \quad \text{and} \quad k_1 + 6 = 0 \]
- Solving these equations gives:
\[ k_2 = -\frac{1}{2} \quad \text{and} \quad k_1 = -6 \]

5. **Finding \( a + 2b + 6c \)**:
- Now we can substitute \( k_1 \) and \( k_2 \) back into our expressions. We need to find \( a + 2b + 6c \):
\[ a + 2b + 6c = a + 2(k_2 a - 3c) + 6c \]
Substituting \( k_2 = -\frac{1}{2} \):
\[ = a + 2\left(-\frac{1}{2}a - 3c\right) + 6c \]
Simplifying this:
\[ = a - a - 6c + 6c = 0 \]

### Conclusion:
Thus, we find that:
\[ a + 2b + 6c = 0 \]