Let O be the origin and` vec(OX) , vec(OY) , vec(OZ)` be three unit vector in the directions of the sides `vec(QR) , vec(RP),vec(PQ)` respectively , of a triangle PQR.
`|vec(OX)xxvec(OY)|=`

Options

A. sin (P + Q)
B. sin 2R
C. sin (P+R)
D. sin (Q+R)

Explanation

To solve the problem, we need to find the modulus of the cross product of the unit vectors \(\vec{OX}\) and \(\vec{OY}\) that are in the directions of the sides of triangle \(PQR\).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Let \(\vec{OX}\) be the unit vector in the direction of side \(QR\).
- Let \(\vec{OY}\) be the unit vector in the direction of side \(RP\).
- Since these are unit vectors, we can express them as:
\[ \vec{OX} = \frac{\vec{QR}}{|\vec{QR}|}, \quad \vec{OY} = \frac{\vec{RP}}{|\vec{RP}|} \]

2. **Finding the Angle Between the Vectors**:
- The angle \(\theta\) between \(\vec{OX}\) and \(\vec{OY}\) can be determined using the properties of the triangle. The angle between the two vectors corresponds to the exterior angle at vertex \(R\) of triangle \(PQR\).
- By the exterior angle theorem, we have:
\[ \theta = \angle P + \angle Q \]

3. **Calculating the Cross Product**:
- The modulus of the cross product of two vectors is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\phi) \]
where \(\phi\) is the angle between the vectors \(\vec{a}\) and \(\vec{b}\).
- Applying this to our unit vectors:
\[ |\vec{OX} \times \vec{OY}| = |\vec{OX}| |\vec{OY}| \sin(\theta) \]
- Since \(\vec{OX}\) and \(\vec{OY}\) are unit vectors, we have:
\[ |\vec{OX}| = 1, \quad |\vec{OY}| = 1 \]
- Therefore:
\[ |\vec{OX} \times \vec{OY}| = 1 \cdot 1 \cdot \sin(\theta) = \sin(\theta) \]

4. **Substituting the Angle**:
- We substitute \(\theta\) with \(\angle P + \angle Q\):
\[ |\vec{OX} \times \vec{OY}| = \sin(\angle P + \angle Q) \]

5. **Final Result**:
- Thus, the modulus of the cross product of the unit vectors \(\vec{OX}\) and \(\vec{OY}\) is:
\[ |\vec{OX} \times \vec{OY}| = \sin(P + Q) \]

### Conclusion:
The final answer is:
\[ |\vec{OX} \times \vec{OY}| = \sin(P + Q) \]