In a `DeltaABC`, if 2AC=3CB, then 2OA+3OB is equal to

Options

A. 5OC
B. `-OC`
C. `OC`
D. none of these

Explanation

To solve the problem, we need to find the value of \(2OA + 3OB\) given that \(2AC = 3CB\) in triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Given Condition:**
We are given that \(2AC = 3CB\). This implies a relationship between the segments \(AC\) and \(CB\).

2. **Setting Up the Triangle:**
Let's denote the points of the triangle as \(A\), \(B\), and \(C\). We will also introduce a point \(O\) outside the triangle.

3. **Expressing Vectors:**
We can express the vectors \(OA\) and \(OB\) in terms of \(OC\):
- In triangle \(OAC\):
\[ OA = OC + CA \]
- In triangle \(OBC\):
\[ OB = OC + CB \]

4. **Substituting into the Expression:**
Now, we substitute these expressions into \(2OA + 3OB\):
\[ 2OA + 3OB = 2(OC + CA) + 3(OC + CB) \]

5. **Distributing the Scalars:**
Distributing the scalars gives us:
\[ = 2OC + 2CA + 3OC + 3CB \]

6. **Combining Like Terms:**
Now, combine the terms:
\[ = (2OC + 3OC) + 2CA + 3CB = 5OC + 2CA + 3CB \]

7. **Using the Given Condition:**
From the given condition \(2AC = 3CB\), we can express \(AC\) in terms of \(CB\):
- Rearranging gives us \(CA = \frac{3}{2}CB\).
- Therefore, substituting \(CA\) into our equation:
\[ 2CA = 2 \left(\frac{3}{2}CB\right) = 3CB \]

8. **Substituting Back:**
Now substitute \(2CA\) back into our expression:
\[ 5OC + 3CB + 3CB = 5OC + 6CB \]

9. **Final Expression:**
Since \(2AC = 3CB\), we can conclude that \(2CA + 3CB = 0\) (as they are equal in magnitude but opposite in direction). Thus:
\[ 2CA + 3CB = 0 \Rightarrow 5OC + 0 = 5OC \]

### Conclusion:
Thus, the final result is:
\[ 2OA + 3OB = 5OC \]