Suppose a, b, c are three unit vectors such that
`" "abs(a-b)^(2)+abs(b-c)^(2)+abs(c-a)^(2)=9`,
then `abs(2a+7b+7c)=`___________

Answer

To solve the problem step by step, we need to find the value of \( |2a + 7b + 7c| \) given that \( |a - b|^2 + |b - c|^2 + |c - a|^2 = 9 \) and \( a, b, c \) are unit vectors.

### Step 1: Expand the given equation

We start with the equation:

\[ |a - b|^2 + |b - c|^2 + |c - a|^2 = 9 \]

Using the property of dot products, we can expand each term:

\[ |a - b|^2 = (a - b) \cdot (a - b) = |a|^2 - 2a \cdot b + |b|^2 \]

Since \( a \) and \( b \) are unit vectors, \( |a|^2 = 1 \) and \( |b|^2 = 1 \). Thus, we have:

\[ |a - b|^2 = 1 - 2a \cdot b + 1 = 2 - 2a \cdot b \]

Similarly, we can expand the other two terms:

\[ |b - c|^2 = 2 - 2b \cdot c \]
\[ |c - a|^2 = 2 - 2c \cdot a \]

### Step 2: Combine the expanded terms

Now we can combine these results:

\[ (2 - 2a \cdot b) + (2 - 2b \cdot c) + (2 - 2c \cdot a) = 9 \]

This simplifies to:

\[ 6 - 2(a \cdot b + b \cdot c + c \cdot a) = 9 \]

### Step 3: Solve for the dot products

Rearranging gives us:

\[ -2(a \cdot b + b \cdot c + c \cdot a) = 3 \]

Dividing by -2:

\[ a \cdot b + b \cdot c + c \cdot a = -\frac{3}{2} \]

### Step 4: Use the relationship of the vectors

Since \( a + b + c = 0 \) (derived from the condition that the sum of the squares is zero), we can express \( b + c \) as:

\[ b + c = -a \]

### Step 5: Substitute into the expression we need to find

Now we substitute \( b + c \) into our expression:

\[ |2a + 7b + 7c| = |2a + 7(-a)| = |2a - 7a| = |-5a| = 5|a| \]

Since \( a \) is a unit vector, \( |a| = 1 \):

\[ |2a + 7b + 7c| = 5 \cdot 1 = 5 \]

### Final Answer

Thus, the value of \( |2a + 7b + 7c| \) is:

\[ \boxed{5} \]