Orthocenter of an equilateral triangle ABC is the origin O. If `vec(OA)=veca, vec(OB)=vecb, vec(OC)=vecc`, then `vec(AB)+2vec(BC)+3vec(CA)=`

Options

A. `3vecc`
B. `3veca`
C. `vec0`
D. `3vecb`

Explanation

To solve the problem, we need to find the value of the expression \( \vec{AB} + 2\vec{BC} + 3\vec{CA} \) given that the orthocenter of triangle \( ABC \) is the origin \( O \) and the vectors from the origin to the vertices are defined as \( \vec{OA} = \vec{a} \), \( \vec{OB} = \vec{b} \), and \( \vec{OC} = \vec{c} \).

### Step-by-Step Solution:

1. **Express the vectors in terms of \( \vec{a}, \vec{b}, \vec{c} \)**:
- The vector \( \vec{AB} \) can be expressed as:
\[ \vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a} \]

2. **Calculate \( \vec{BC} \)**:
- The vector \( \vec{BC} \) can be expressed as:
\[ \vec{BC} = \vec{OC} - \vec{OB} = \vec{c} - \vec{b} \]

3. **Calculate \( \vec{CA} \)**:
- The vector \( \vec{CA} \) can be expressed as:
\[ \vec{CA} = \vec{OA} - \vec{OC} = \vec{a} - \vec{c} \]

4. **Substitute these expressions into the original equation**:
- Now substitute \( \vec{AB}, \vec{BC}, \vec{CA} \) into the expression \( \vec{AB} + 2\vec{BC} + 3\vec{CA} \):
\[ \vec{AB} + 2\vec{BC} + 3\vec{CA} = (\vec{b} - \vec{a}) + 2(\vec{c} - \vec{b}) + 3(\vec{a} - \vec{c}) \]

5. **Distribute and combine like terms**:
- Expanding the expression gives:
\[ = \vec{b} - \vec{a} + 2\vec{c} - 2\vec{b} + 3\vec{a} - 3\vec{c} \]
- Combine the terms:
\[ = (-\vec{a} + 3\vec{a}) + (\vec{b} - 2\vec{b}) + (2\vec{c} - 3\vec{c}) \]
\[ = 2\vec{a} - \vec{b} - \vec{c} \]

6. **Use the property of the orthocenter**:
- Since \( O \) is the orthocenter of triangle \( ABC \), we have:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
- Therefore, \( \vec{c} = -(\vec{a} + \vec{b}) \).

7. **Substituting back into the expression**:
- Substitute \( \vec{c} \) into \( 2\vec{a} - \vec{b} - \vec{c} \):
\[ = 2\vec{a} - \vec{b} - (-\vec{a} - \vec{b}) \]
\[ = 2\vec{a} - \vec{b} + \vec{a} + \vec{b} \]
\[ = 3\vec{a} \]

### Final Result:
Thus, the final result is:
\[ \vec{AB} + 2\vec{BC} + 3\vec{CA} = 3\vec{a} \]