If a, b and c are unit vectors satisfying `|a-b|^(2)+|b-c|^(2)+|c-a|^(2)=9`, then `|2a+5b+5x|` is

Answer

To solve the problem, we need to find the value of \( |2a + 5b + 5c| \) given that \( |a - b|^2 + |b - c|^2 + |c - a|^2 = 9 \) and that \( a, b, c \) are unit vectors.

### Step-by-Step Solution:

1. **Understanding the given equation**:
We start with the equation:
\[ |a - b|^2 + |b - c|^2 + |c - a|^2 = 9 \]
Since \( a, b, c \) are unit vectors, we know that \( |a|^2 = |b|^2 = |c|^2 = 1 \).

2. **Expanding the squares**:
We can expand each term:
\[ |a - b|^2 = |a|^2 + |b|^2 - 2a \cdot b = 1 + 1 - 2a \cdot b = 2 - 2a \cdot b \]
\[ |b - c|^2 = |b|^2 + |c|^2 - 2b \cdot c = 1 + 1 - 2b \cdot c = 2 - 2b \cdot c \]
\[ |c - a|^2 = |c|^2 + |a|^2 - 2c \cdot a = 1 + 1 - 2c \cdot a = 2 - 2c \cdot a \]

3. **Substituting back into the equation**:
Now substituting these back into the original equation:
\[ (2 - 2a \cdot b) + (2 - 2b \cdot c) + (2 - 2c \cdot a) = 9 \]
Simplifying this gives:
\[ 6 - 2(a \cdot b + b \cdot c + c \cdot a) = 9 \]
Rearranging gives:
\[ -2(a \cdot b + b \cdot c + c \cdot a) = 3 \]
Thus:
\[ a \cdot b + b \cdot c + c \cdot a = -\frac{3}{2} \]

4. **Finding \( |2a + 5b + 5c| \)**:
We need to find \( |2a + 5b + 5c| \). We can calculate this using the formula for the magnitude of a vector:
\[ |2a + 5b + 5c|^2 = |2a|^2 + |5b|^2 + |5c|^2 + 2(2a \cdot 5b + 2a \cdot 5c + 5b \cdot 5c) \]
This simplifies to:
\[ |2a + 5b + 5c|^2 = 4|a|^2 + 25|b|^2 + 25|c|^2 + 20(a \cdot b + a \cdot c + b \cdot c) \]
Since \( |a|^2 = |b|^2 = |c|^2 = 1 \):
\[ |2a + 5b + 5c|^2 = 4 + 25 + 25 + 20(-\frac{3}{2}) \]
\[ = 54 - 30 = 24 \]

5. **Taking the square root**:
Finally, we take the square root to find the magnitude:
\[ |2a + 5b + 5c| = \sqrt{24} = 2\sqrt{6} \]

### Final Answer:
Thus, the value of \( |2a + 5b + 5c| \) is \( 2\sqrt{6} \).