If S is the cirucmcentre, G the centroid, O the orthocentre of a triangle ABC, then `vec(SA) + vec(SB) + vec(SC)` is:

Options

A. `3vec(SG)`
B. `vec(OS)`
C. `vec(SO)`
D. `vec(OG)`

Answer

To solve the problem, we need to find the value of the vector sum \( \vec{SA} + \vec{SB} + \vec{SC} \) where \( S \) is the circumcenter, \( G \) is the centroid, and \( O \) is the orthocenter of triangle \( ABC \).

### Step-by-step Solution:

1. **Express the vectors in terms of the origin:**
- We know that the vector \( \vec{SA} \) can be expressed as:
\[ \vec{SA} = \vec{OA} - \vec{OS} \]
- Similarly, we can express \( \vec{SB} \) and \( \vec{SC} \):
\[ \vec{SB} = \vec{OB} - \vec{OS} \]
\[ \vec{SC} = \vec{OC} - \vec{OS} \]

2. **Combine the vectors:**
- Now, we can add these three vectors together:
\[ \vec{SA} + \vec{SB} + \vec{SC} = (\vec{OA} - \vec{OS}) + (\vec{OB} - \vec{OS}) + (\vec{OC} - \vec{OS}) \]
- This simplifies to:
\[ = \vec{OA} + \vec{OB} + \vec{OC} - 3\vec{OS} \]

3. **Recognize the centroid:**
- The centroid \( G \) of triangle \( ABC \) is given by:
\[ \vec{G} = \frac{\vec{OA} + \vec{OB} + \vec{OC}}{3} \]
- Therefore, we can substitute this into our equation:
\[ \vec{SA} + \vec{SB} + \vec{SC} = 3\vec{G} - 3\vec{OS} \]
- Factoring out the 3 gives us:
\[ = 3\left(\vec{G} - \vec{S}\right) \]

4. **Relate \( \vec{G} \) and \( \vec{S} \):**
- We know from the properties of triangle centers that the vector \( \vec{SG} \) (from circumcenter to centroid) can be expressed in terms of the orthocenter \( O \):
\[ \vec{SG} = \frac{1}{3} \vec{SO} \]
- Thus, we can write:
\[ \vec{G} - \vec{S} = \frac{1}{3} \vec{SO} \]

5. **Final expression:**
- Substituting this back into our equation gives:
\[ \vec{SA} + \vec{SB} + \vec{SC} = 3\left(\frac{1}{3} \vec{SO}\right) = \vec{SO} \]

### Conclusion:
Thus, we conclude that:
\[ \vec{SA} + \vec{SB} + \vec{SC} = \vec{SO} \]