Let ` veca , vecb,vecc` be three vectors such that
` veca bot ( vecb + vecc), vecb bot ( vecc + veca) and vecc bot ( veca + vecb) , " if " |veca| =1 , |vecb| =2 `,
` |vecc| =3 , " then " | veca + vecb + vecc|` is,

Options

A. ` sqrt6`
B. 14
C. ` sqrt14`
D. none of these

Explanation

To solve the problem, we need to find the magnitude of the vector sum \( \vec{a} + \vec{b} + \vec{c} \) given the conditions of perpendicularity and the magnitudes of the vectors.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have three vectors \( \vec{a}, \vec{b}, \vec{c} \).
- The conditions given are:
- \( \vec{a} \perp (\vec{b} + \vec{c}) \) implies \( \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \).
- \( \vec{b} \perp (\vec{c} + \vec{a}) \) implies \( \vec{b} \cdot (\vec{c} + \vec{a}) = 0 \).
- \( \vec{c} \perp (\vec{a} + \vec{b}) \) implies \( \vec{c} \cdot (\vec{a} + \vec{b}) = 0 \).
- The magnitudes are given as \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), and \( |\vec{c}| = 3 \).

2. **Set Up the Dot Product Equations**:
- From \( \vec{a} \cdot (\vec{b} + \vec{c}) = 0 \):
\[ \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \quad (1) \]
- From \( \vec{b} \cdot (\vec{c} + \vec{a}) = 0 \):
\[ \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \quad (2) \]
- From \( \vec{c} \cdot (\vec{a} + \vec{b}) = 0 \):
\[ \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \quad (3) \]

3. **Identify Zero Dot Products**:
- From equations (1), (2), and (3), we can conclude:
- \( \vec{a} \cdot \vec{b} = -\vec{a} \cdot \vec{c} \)
- \( \vec{b} \cdot \vec{c} = -\vec{b} \cdot \vec{a} \)
- \( \vec{c} \cdot \vec{a} = -\vec{c} \cdot \vec{b} \)
- This means that all dot products between different vectors are zero:
\[ \vec{a} \cdot \vec{b} = 0, \quad \vec{b} \cdot \vec{c} = 0, \quad \vec{c} \cdot \vec{a} = 0 \]

4. **Calculate the Magnitude of \( \vec{a} + \vec{b} + \vec{c} \)**:
- Let \( \vec{x} = \vec{a} + \vec{b} + \vec{c} \).
- The magnitude squared is given by:
\[ |\vec{x}|^2 = |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
- Since all dot products are zero:
\[ |\vec{x}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 \]
- Substitute the magnitudes:
\[ |\vec{x}|^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \]
- Therefore:
\[ |\vec{x}| = \sqrt{14} \]

5. **Final Answer**:
- The magnitude of \( \vec{a} + \vec{b} + \vec{c} \) is \( \sqrt{14} \).