In a triangle `PQR, "let: aveca= vecQR= vecb= vecRP and vec c= vecPQ`. ,
If `|veca|=3, |vecb|=4 and (veca. (vec c. vecb))/(vec. (veca- vecb))=(|veca|)/(|veca|+|vecb|)` then the value of `|vecaxx vecb|^(2)` is _____

Answer

To solve the problem, we start with the given vectors in the triangle \( PQR \):

1. **Define the vectors**:
- Let \( \vec{a} = \vec{RP} \)
- Let \( \vec{b} = \vec{QR} \)
- Let \( \vec{c} = \vec{PQ} \)

We know that:
- \( |\vec{a}| = 3 \)
- \( |\vec{b}| = 4 \)

2. **Use the given equation**:
The problem states:
\[ \frac{\vec{a} \cdot (\vec{c} - \vec{b})}{\vec{c} \cdot (\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|} \]
Substituting the magnitudes:
\[ \frac{\vec{a} \cdot (\vec{c} - \vec{b})}{\vec{c} \cdot (\vec{a} - \vec{b})} = \frac{3}{3 + 4} = \frac{3}{7} \]

3. **Express \( \vec{c} \)**:
From the triangle property, we know:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \implies \vec{c} = -(\vec{a} + \vec{b}) \]

4. **Substitute \( \vec{c} \)**:
Substitute \( \vec{c} \) into the equation:
\[ \vec{a} \cdot (-\vec{a}) = -|\vec{a}|^2 = -9 \]
\[ \vec{c} \cdot (\vec{a} - \vec{b}) = -(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = -(\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b}) = -(|\vec{a}|^2 - \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{b} - |\vec{b}|^2) \]
This simplifies to:
\[ -(|\vec{a}|^2 - |\vec{b}|^2) = -9 + 16 = 7 \]

5. **Set up the equation**:
Now we have:
\[ \frac{-9}{7} = \frac{3}{7} \]
This is consistent with our earlier substitution.

6. **Find \( |\vec{a} \times \vec{b}|^2 \)**:
We know from the properties of vectors:
\[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \]
We already have:
- \( |\vec{a}|^2 = 9 \)
- \( |\vec{b}|^2 = 16 \)
- \( \vec{a} \cdot \vec{b} = -6 \) (as derived from previous steps)

Thus:
\[ |\vec{a} \times \vec{b}|^2 + (-6)^2 = 9 \times 16 \]
\[ |\vec{a} \times \vec{b}|^2 + 36 = 144 \]
\[ |\vec{a} \times \vec{b}|^2 = 144 - 36 = 108 \]

7. **Final answer**:
The value of \( |\vec{a} \times \vec{b}|^2 \) is \( \boxed{108} \).