Let `vec(a), vec(b), vec(c)` be vectors of length `3, 4, 5` respectively. Let `vec(a)` be perpendicular to `vec(b)+vec(c), vec(b)` to `vec(c)+vec(a)` and `vec(c)` to `vec(a)+vec(b)`. Then `|vec(a)+vec(b)+vec(c)|` is :

Options

A. `2sqrt(5)`
B. `2sqrt(2)`
C. `10sqrt(5)`
D. `5sqrt(2)`

Answer

To solve the problem step by step, we will analyze the given conditions and derive the required modulus of the sum of the vectors.

### Step 1: Understand the conditions
We have three vectors \( \vec{a}, \vec{b}, \vec{c} \) with lengths 3, 4, and 5 respectively. The conditions state that:
- \( \vec{a} \) is perpendicular to \( \vec{b} + \vec{c} \)
- \( \vec{b} \) is perpendicular to \( \vec{c} + \vec{a} \)
- \( \vec{c} \) is perpendicular to \( \vec{a} + \vec{b} \)

### Step 2: Set up the equations
From the perpendicularity conditions, we can write:
1. \( \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \) implies \( \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \) (Equation 1)
2. \( \vec{b} \cdot (\vec{c} + \vec{a}) = 0 \) implies \( \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \) (Equation 2)
3. \( \vec{c} \cdot (\vec{a} + \vec{b}) = 0 \) implies \( \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \) (Equation 3)

### Step 3: Add the equations
Adding all three equations:
\[ (\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) + (\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a}) + (\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b}) = 0 \]
This simplifies to:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
Thus, we have:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0 \quad \text{(Equation 4)} \]

### Step 4: Calculate the modulus of \( \vec{a} + \vec{b} + \vec{c} \)
The modulus squared of the sum of the vectors is given by:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Substituting the lengths of the vectors:
- \( |\vec{a}|^2 = 3^2 = 9 \)
- \( |\vec{b}|^2 = 4^2 = 16 \)
- \( |\vec{c}|^2 = 5^2 = 25 \)

Thus:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 9 + 16 + 25 + 2(0) = 50 \]

### Step 5: Find the modulus
Taking the square root gives:
\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2} \]

### Conclusion
Therefore, the modulus of \( \vec{a} + \vec{b} + \vec{c} \) is \( 5\sqrt{2} \).