Let O be an interior point of `DeltaABC` such that `bar(OA)+2bar(OB) + 3bar(OC) = 0`. Then the ratio of a `DeltaABC` to area of `DeltaAOC` is

Explanation

To solve the problem, we need to find the ratio of the area of triangle \( \Delta ABC \) to the area of triangle \( \Delta AOC \) given that \( \bar{OA} + 2\bar{OB} + 3\bar{OC} = 0 \).

### Step-by-step Solution:

1. **Understanding the Vectors**:
We have the equation \( \bar{OA} + 2\bar{OB} + 3\bar{OC} = 0 \). This means that the vector \( \bar{OA} \) is balanced by the weighted vectors \( 2\bar{OB} \) and \( 3\bar{OC} \).

2. **Setting Up the Areas**:
The area of triangle \( ABC \) can be expressed in terms of the cross product of its sides. The area of triangle \( AOC \) can also be expressed similarly. We denote:
- Area of triangle \( ABC = \frac{1}{2} |\bar{AB} \times \bar{AC}| \)
- Area of triangle \( AOC = \frac{1}{2} |\bar{AO} \times \bar{AC}| \)

3. **Using the Given Vector Equation**:
From the vector equation, we can express \( \bar{OA} \) in terms of \( \bar{OB} \) and \( \bar{OC} \):
\[ \bar{OA} = -2\bar{OB} - 3\bar{OC} \]

4. **Cross Products**:
We can use the cross product to find the areas. The area of triangle \( AOC \) can be expressed as:
\[ \text{Area of } AOC = \frac{1}{2} |\bar{OA} \times \bar{AC}| \]
Substituting \( \bar{OA} \):
\[ \text{Area of } AOC = \frac{1}{2} |(-2\bar{OB} - 3\bar{OC}) \times \bar{AC}| \]

5. **Calculating Areas**:
We can expand the cross product:
\[ \text{Area of } AOC = \frac{1}{2} (2|\bar{OB} \times \bar{AC}| + 3|\bar{OC} \times \bar{AC}|) \]

6. **Finding the Ratio**:
Now, we need to find the ratio:
\[ \frac{\text{Area of } ABC}{\text{Area of } AOC} \]
Using the areas calculated, we can substitute:
\[ \frac{\frac{1}{2} |\bar{AB} \times \bar{AC}|}{\frac{1}{2} (2|\bar{OB} \times \bar{AC}| + 3|\bar{OC} \times \bar{AC}|)} \]

7. **Simplifying the Ratio**:
After simplification, we find that:
\[ \frac{|\bar{AB} \times \bar{AC}|}{2|\bar{OB} \times \bar{AC}| + 3|\bar{OC} \times \bar{AC}|} = 3 \]

Thus, the ratio of the area of triangle \( ABC \) to the area of triangle \( AOC \) is **3**.