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Step 1: Interpret the Given Information
We have three non-zero vectors $ \vec{a}, \vec{b}, \vec{c} $ that are pairwise non-collinear. The statements given are:
$ \vec{a} + 3\vec{b} $ is collinear with $ \vec{c} $.
$ \vec{b} + 2\vec{c} $ is collinear with $ \vec{a} $.
We need to find the vector $ \vec{a} + \vec{b} + 6\vec{c} $.
Step 2: Express Collinearity Conditions
Since $ \vec{a} + 3\vec{b} $ is collinear with $ \vec{c} $, there exists a scalar $ \lambda $ such that
$ \vec{a} + 3\vec{b} = \lambda \, \vec{c} \quad \ldots (1)$
Since $ \vec{b} + 2\vec{c} $ is collinear with $ \vec{a} $, there exists a scalar $ \mu $ such that
$ \vec{b} + 2\vec{c} = \mu \, \vec{a} \quad \ldots (2)$
Step 3: Relate to the Desired Expression
We aim to find $ \vec{a} + \vec{b} + 6\vec{c} $. Notice we can form this expression by suitably adding terms to equations (1) and (2) and then comparing results:
From equation (1):
$ \vec{a} + 3\vec{b} = \lambda \, \vec{c} $
Add $ 6\vec{c} $ to both sides to get:
$ \vec{a} + 3\vec{b} + 6\vec{c} = \lambda \, \vec{c} + 6\vec{c} = (\lambda + 6)\vec{c} \quad \ldots (3)$
From equation (2):
$ \vec{b} + 2\vec{c} = \mu \, \vec{a} $
Multiply both sides by 3:
$ 3\vec{b} + 6\vec{c} = 3\mu \, \vec{a} \quad \ldots (4)$
Add $ \vec{a} $ to both sides of (4):
$ \vec{a} + 3\vec{b} + 6\vec{c} = \vec{a} + 3\mu \, \vec{a} = (1 + 3\mu)\vec{a} \quad \ldots (5)$
Step 4: Equate the Two Forms
Equations (3) and (5) represent the same vector $ \vec{a} + 3\vec{b} + 6\vec{c} $, but in different directions ($ \vec{c} $ vs. $ \vec{a} $). Hence,
$ (\lambda + 6)\vec{c} = (1 + 3\mu)\vec{a} \quad \ldots (6)$
Because $ \vec{a} $ and $ \vec{c} $ are not collinear, the only way for this equality to hold is if both scalar coefficients are zero:
$ \lambda + 6 = 0 \quad \text{and} \quad 1 + 3\mu = 0.$
Thus,
$ \lambda = -6, \quad \mu = -\frac{1}{3}.$
Step 5: Conclude the Value of $ \vec{a} + \vec{b} + 6\vec{c} $
Using $ \lambda = -6 $ in equation (3) gives
$ \vec{a} + 3\vec{b} + 6\vec{c} = (\lambda + 6)\vec{c} = ( -6 + 6 )\vec{c} = \vec{0}.$
Hence,
$ \displaystyle \vec{a} + \vec{b} + 6\vec{c} = \vec{0}.$
Final Answer
The vector $ \vec{a} + \vec{b} + 6\vec{c} $ is $ \vec{0} $.
