In a `Delta ABC, AD:DB=1:3, AE:EC=2:3, BF:FC=1:2`. Find the ratio of area of `Delta` of `Delta DEF: Delta ABC`.

Answer

To find the ratio of the area of triangle DEF to triangle ABC given the ratios of segments AD:DB, AE:EC, and BF:FC, we will follow these steps:

### Step 1: Understand the Ratios
We have the following ratios:
- \( AD:DB = 1:3 \)
- \( AE:EC = 2:3 \)
- \( BF:FC = 1:2 \)

### Step 2: Assign Lengths to Segments
Let:
- \( AD = 1x \) and \( DB = 3x \) (thus, \( AB = AD + DB = 4x \))
- \( AE = 2y \) and \( EC = 3y \) (thus, \( AC = AE + EC = 5y \))
- \( BF = 1z \) and \( FC = 2z \) (thus, \( BC = BF + FC = 3z \))

### Step 3: Calculate the Area of Triangle ABC
The area of triangle ABC can be calculated using the formula:
\[ \text{Area}_{ABC} = \frac{1}{2} \times AB \times AC \times \sin(B) \]
Substituting the lengths:
\[ \text{Area}_{ABC} = \frac{1}{2} \times (4x) \times (5y) \times \sin(B) = 10xy \sin(B) \]

### Step 4: Calculate the Area of Triangle DEF
To find the area of triangle DEF, we will use the ratios of the segments to find the areas of triangles ADB, AEC, and BFC.

1. **Area of Triangle ADB**:
\[ \text{Area}_{ADB} = \frac{1}{2} \times AD \times AB \times \sin(B) = \frac{1}{2} \times (1x) \times (4x) \times \sin(B) = 2x^2 \sin(B) \]

2. **Area of Triangle AEC**:
\[ \text{Area}_{AEC} = \frac{1}{2} \times AE \times AC \times \sin(B) = \frac{1}{2} \times (2y) \times (5y) \times \sin(B) = 5y^2 \sin(B) \]

3. **Area of Triangle BFC**:
\[ \text{Area}_{BFC} = \frac{1}{2} \times BF \times BC \times \sin(B) = \frac{1}{2} \times (1z) \times (3z) \times \sin(B) = \frac{3}{2} z^2 \sin(B) \]

### Step 5: Calculate the Area of Triangle DEF
The area of triangle DEF can be found by subtracting the areas of triangles ADB, AEC, and BFC from the area of triangle ABC:
\[ \text{Area}_{DEF} = \text{Area}_{ABC} - (\text{Area}_{ADB} + \text{Area}_{AEC} + \text{Area}_{BFC}) \]
Substituting the areas:
\[ \text{Area}_{DEF} = 10xy \sin(B) - (2x^2 \sin(B) + 5y^2 \sin(B) + \frac{3}{2} z^2 \sin(B)) \]
Factoring out \(\sin(B)\):
\[ \text{Area}_{DEF} = \sin(B) \left(10xy - (2x^2 + 5y^2 + \frac{3}{2} z^2)\right) \]

### Step 6: Find the Ratio of Areas
Now, we can find the ratio of the areas:
\[ \text{Ratio} = \frac{\text{Area}_{DEF}}{\text{Area}_{ABC}} = \frac{\sin(B) \left(10xy - (2x^2 + 5y^2 + \frac{3}{2} z^2)\right)}{10xy \sin(B)} \]
Cancelling \(\sin(B)\):
\[ \text{Ratio} = \frac{10xy - (2x^2 + 5y^2 + \frac{3}{2} z^2)}{10xy} \]

### Final Ratio
To get the final ratio, we need to evaluate the expression based on the values of \(x\), \(y\), and \(z\). However, we can conclude that the ratio of the areas of triangle DEF to triangle ABC is a function of the segments defined by the ratios provided.