`|vec(a)|=2,|vec(b)|=3,|vec(c)|=6` . Angle between `vec(a)` and `vec(b),vec(b)` and `vec(c)` and `vec(c)` and `vec(a)` is `120^(@)` each, find `|vec(a)+vec(b)+vec(c)|`?

Explanation

To find the magnitude of the vector sum \(|\vec{a} + \vec{b} + \vec{c}|\), we can use the formula for the magnitude of the sum of vectors:

\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})} \]

### Step 1: Calculate the magnitudes squared
Given:
- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 3\)
- \(|\vec{c}| = 6\)

We calculate:
\[ |\vec{a}|^2 = 2^2 = 4 \]
\[ |\vec{b}|^2 = 3^2 = 9 \]
\[ |\vec{c}|^2 = 6^2 = 36 \]

### Step 2: Sum the squares of the magnitudes
Now we can sum these values:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 4 + 9 + 36 = 49 \]

### Step 3: Calculate the dot products
Next, we need to calculate the dot products. The angle between each pair of vectors is \(120^\circ\), and we know:
\[ \cos(120^\circ) = -\frac{1}{2} \]

Now we calculate each dot product:
1. \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(120^\circ) = 2 \cdot 3 \cdot \left(-\frac{1}{2}\right) = -3\)
2. \(\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(120^\circ) = 3 \cdot 6 \cdot \left(-\frac{1}{2}\right) = -9\)
3. \(\vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos(120^\circ) = 6 \cdot 2 \cdot \left(-\frac{1}{2}\right) = -6\)

### Step 4: Sum the dot products
Now we sum the dot products:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3 - 9 - 6 = -18 \]

### Step 5: Substitute into the magnitude formula
Now we substitute everything back into the formula:
\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{49 + 2(-18)} \]
\[ = \sqrt{49 - 36} = \sqrt{13} \]

### Final Answer
Thus, the magnitude of \(|\vec{a} + \vec{b} + \vec{c}|\) is:
\[ \sqrt{13} \]