`veca, vecb ,vecc` are three non zero vectors no two of which are collonear and the vectors `veca + vecb` be collinear with `vecc,vecb+vecc` to collinear with `veca` then `veca+vecb+vecc` the equal to ? (A) `veca` (B) `vecb` (C) `vecc` (D) None of these

Options

A. `vec a `
B. `vec b`
C. `vec c`
D. `vec 0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the conditions given in the problem
We have three non-zero vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
1. \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \).
2. \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \).

### Step 2: Express the collinearity conditions mathematically
From the first condition, since \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \), we can write:
\[ \vec{a} + \vec{b} = \lambda \vec{c} \quad (1) \]
for some scalar \( \lambda \).

From the second condition, since \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \), we can write:
\[ \vec{b} + \vec{c} = \mu \vec{a} \quad (2) \]
for some scalar \( \mu \).

### Step 3: Substitute and manipulate the equations
From equation (1), we can express \( \vec{c} \) as:
\[ \vec{c} = \frac{\vec{a} + \vec{b}}{\lambda} \quad (3) \]

From equation (2), we can express \( \vec{a} \) as:
\[ \vec{a} = \frac{\vec{b} + \vec{c}}{\mu} \quad (4) \]

### Step 4: Substitute equation (3) into equation (4)
Substituting equation (3) into equation (4):
\[ \vec{a} = \frac{\vec{b} + \frac{\vec{a} + \vec{b}}{\lambda}}{\mu} \]
This simplifies to:
\[ \vec{a} = \frac{\lambda \vec{b} + \vec{a} + \vec{b}}{\lambda \mu} \]
Rearranging gives:
\[ \lambda \mu \vec{a} = \lambda \vec{b} + \vec{a} + \vec{b} \]
This can be further simplified to:
\[ (\lambda \mu - 1) \vec{a} = (\lambda + 1) \vec{b} \]

### Step 5: Analyze the coefficients
Since \( \vec{a} \) and \( \vec{b} \) are non-collinear, we can set the coefficients equal to zero:
1. \( \lambda \mu - 1 = 0 \)
2. \( \lambda + 1 = 0 \)

From \( \lambda + 1 = 0 \), we find:
\[ \lambda = -1 \]
Substituting \( \lambda = -1 \) into \( \lambda \mu - 1 = 0 \):
\[ -1 \cdot \mu - 1 = 0 \implies -\mu - 1 = 0 \implies \mu = -1 \]

### Step 6: Substitute back to find \( \vec{a} + \vec{b} + \vec{c} \)
Now substituting \( \lambda = -1 \) back into equation (1):
\[ \vec{a} + \vec{b} = -\vec{c} \]
Thus:
\[ \vec{c} = -(\vec{a} + \vec{b}) \]

Now substituting into \( \vec{a} + \vec{b} + \vec{c} \):
\[ \vec{a} + \vec{b} + \vec{c} = \vec{a} + \vec{b} - (\vec{a} + \vec{b}) = 0 \]

### Final Result
Thus, we conclude that:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \]

### Answer
The answer is (D) None of these, since the result is the zero vector.