Given that the vectors `bara,barb and barc` (no two of them are collinera). Further if `(bara +barb)` is collinear with `barc,(barb+barc)` is collinear with `bara and |bara|=|barb|=|barc|=sqrt2`. Then the value of `|bara.barb+barb.barc+barc.bara|` is________

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the conditions given
We have three vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
1. No two of them are collinear.
2. \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \).
3. \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \).
4. The magnitudes of all three vectors are equal: \( |\vec{a}| = |\vec{b}| = |\vec{c}| = \sqrt{2} \).

### Step 2: Analyze the collinearity conditions
Since \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \), we can express this as:
\[ \vec{c} = k(\vec{a} + \vec{b}) \quad \text{for some scalar } k \]
Similarly, since \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \):
\[ \vec{a} = m(\vec{b} + \vec{c}) \quad \text{for some scalar } m \]

### Step 3: Determine the angles between the vectors
Given that the vectors are not collinear and are of equal magnitude, we can infer that the angle between any two vectors \( \vec{a}, \vec{b}, \vec{c} \) is \( 120^\circ \). This is because the vectors are symmetrically arranged in space.

### Step 4: Calculate the dot products
Using the cosine of the angle between the vectors:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(120^\circ) = \sqrt{2} \cdot \sqrt{2} \cdot \left(-\frac{1}{2}\right) = -1 \]
Similarly, we can calculate:
\[ \vec{b} \cdot \vec{c} = -1 \quad \text{and} \quad \vec{c} \cdot \vec{a} = -1 \]

### Step 5: Sum the dot products
Now we can find the required expression:
\[ |\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}| \]
Substituting the values we found:
\[ |\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}| = |-1 - 1 - 1| = |-3| = 3 \]

### Final Answer
Thus, the value of \( |\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}| \) is \( \boxed{3} \).

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