If `veca, vecb and vecc` are unit vectors satisfying `|veca-vecb|^(2)+|vecb-vecc|^(2)+|vecc-veca|^(2)=9 " then find the value of " |2veca+ 5vecb+ 5vecc|`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have three unit vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \) satisfying the equation:
\[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9 \]

### Step 2: Expand the squares
Using the identity \( |\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y} \), we can expand each term in the equation:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} \]
\[ |\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} \]
\[ |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a} \]

### Step 3: Substitute the unit vector magnitudes
Since \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are unit vectors, we have \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \). Therefore, we can substitute these values:
\[ 1 + 1 - 2\vec{a} \cdot \vec{b} + 1 + 1 - 2\vec{b} \cdot \vec{c} + 1 + 1 - 2\vec{c} \cdot \vec{a} = 9 \]

### Step 4: Simplify the equation
Combining the terms gives:
\[ 3 + 3 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9 \]
This simplifies to:
\[ 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9 \]

### Step 5: Solve for the dot products
Rearranging gives:
\[ -2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3 \]
Thus,
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \]

### Step 6: Find \( |\vec{a} + \vec{b} + \vec{c}|^2 \)
Using the identity \( |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \):
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 2(-\frac{3}{2}) = 3 - 3 = 0 \]
This implies:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \quad \Rightarrow \quad \vec{b} + \vec{c} = -\vec{a} \]

### Step 7: Calculate \( |2\vec{a} + 5\vec{b} + 5\vec{c}| \)
Substituting \( \vec{b} + \vec{c} = -\vec{a} \) into the expression:
\[ |2\vec{a} + 5\vec{b} + 5\vec{c}| = |2\vec{a} + 5(-\vec{a})| = |2\vec{a} - 5\vec{a}| = |-3\vec{a}| \]
Since \( |\vec{a}| = 1 \):
\[ |-3\vec{a}| = 3 \]

### Final Answer
Thus, the value of \( |2\vec{a} + 5\vec{b} + 5\vec{c}| \) is \( \boxed{3} \).