Given a parallelogram `ABCD`. If `|vec(AB)|=a, |vec(AD)| = b & |vec(AC)| = c` , then ` vec(DB) . vec(AB)` has the value

Options

A. `(3a^(2)+b^(2)-c^(2))/(2)`
B. `(a^(2)+3b^(2)-c^(2))/(2)`
C. `(a^(2)-b^(2)+3c^(2))/(2)`
D. None of these

Answer

To solve the problem, we need to find the value of \( \vec{DB} \cdot \vec{AB} \) in the context of the parallelogram \( ABCD \) where \( |\vec{AB}| = a \), \( |\vec{AD}| = b \), and \( |\vec{AC}| = c \).

### Step-by-Step Solution:

1. **Understanding the Parallelogram**:
- We have a parallelogram \( ABCD \) with sides \( \vec{AB} \) and \( \vec{AD} \).
- The diagonals are \( \vec{AC} \) and \( \vec{DB} \).

2. **Using Vector Properties**:
- From the properties of a parallelogram, we know that:
\[ \vec{DB} = \vec{DA} + \vec{AB} \]
- Rearranging gives:
\[ \vec{DA} = \vec{DB} - \vec{AB} \]

3. **Squaring the Equation**:
- Squaring both sides:
\[ |\vec{DA}|^2 = |\vec{DB} - \vec{AB}|^2 \]
- Expanding the right-hand side using the formula \( |\vec{X} - \vec{Y}|^2 = |\vec{X}|^2 + |\vec{Y}|^2 - 2\vec{X} \cdot \vec{Y} \):
\[ |\vec{DA}|^2 = |\vec{DB}|^2 + |\vec{AB}|^2 - 2\vec{DB} \cdot \vec{AB} \]

4. **Substituting Known Values**:
- We know that \( |\vec{AB}| = a \) and \( |\vec{AD}| = b \), so \( |\vec{DA}|^2 = b^2 \).
- Let’s denote \( |\vec{DB}|^2 \) as \( x \):
\[ b^2 = x + a^2 - 2\vec{DB} \cdot \vec{AB} \]

5. **Using the Diagonal Property**:
- The property of the parallelogram states:
\[ 2(|\vec{AB}|^2 + |\vec{AD}|^2) = |\vec{AC}|^2 + |\vec{DB}|^2 \]
- Substituting the known values:
\[ 2(a^2 + b^2) = c^2 + x \]
- Rearranging gives:
\[ x = 2a^2 + 2b^2 - c^2 \]

6. **Substituting Back**:
- Now substitute \( x \) back into the equation for \( b^2 \):
\[ b^2 = (2a^2 + 2b^2 - c^2) + a^2 - 2\vec{DB} \cdot \vec{AB} \]
- Simplifying gives:
\[ b^2 = 3a^2 + b^2 - c^2 - 2\vec{DB} \cdot \vec{AB} \]
- Thus, we can isolate \( \vec{DB} \cdot \vec{AB} \):
\[ 0 = 3a^2 - c^2 - 2\vec{DB} \cdot \vec{AB} \]
- Rearranging gives:
\[ 2\vec{DB} \cdot \vec{AB} = 3a^2 + b^2 - c^2 \]

7. **Final Result**:
- Therefore, we can express \( \vec{DB} \cdot \vec{AB} \) as:
\[ \vec{DB} \cdot \vec{AB} = \frac{3a^2 + b^2 - c^2}{2} \]

### Conclusion:
The value of \( \vec{DB} \cdot \vec{AB} \) is:
\[ \boxed{\frac{3a^2 + b^2 - c^2}{2}} \]