Let ` vec O A= vec a , vec O B=10 vec a+2 vec ba n d vec O C= vec b ,w h e r eO ,Aa n dC` are non-collinear points. Let `p` denotes the areaof quadrilateral `O A C B ,` and let `q` denote the area of parallelogram with `O Aa n dO C` as adjacent sides. If `p=k q ,` then find`kdot`

Explanation

To solve the problem, we need to find the value of \( k \) such that the area of quadrilateral \( OACB \) is \( k \) times the area of the parallelogram formed by vectors \( \vec{OA} \) and \( \vec{OC} \).

### Step-by-Step Solution:

1. **Define the Vectors:**
- Let \( \vec{OA} = \vec{a} \)
- Let \( \vec{OB} = 10\vec{a} + 2\vec{b} \)
- Let \( \vec{OC} = \vec{b} \)

2. **Find the Area of the Parallelogram \( Q \):**
- The area \( Q \) of the parallelogram formed by vectors \( \vec{OA} \) and \( \vec{OC} \) is given by:
\[ Q = |\vec{OA} \times \vec{OC}| = |\vec{a} \times \vec{b}| \]

3. **Find the Area of Quadrilateral \( P \):**
- The area \( P \) of quadrilateral \( OACB \) can be calculated as the sum of the areas of triangles \( OAB \) and \( OBC \).
- The area of triangle \( OAB \) is:
\[ \text{Area of } OAB = \frac{1}{2} |\vec{OA} \times \vec{OB}| = \frac{1}{2} |\vec{a} \times (10\vec{a} + 2\vec{b})| \]
- Using the property of the cross product, we have:
\[ \vec{a} \times (10\vec{a} + 2\vec{b}) = \vec{a} \times 10\vec{a} + \vec{a} \times 2\vec{b} = 0 + 2(\vec{a} \times \vec{b}) = 2(\vec{a} \times \vec{b}) \]
- Thus, the area of triangle \( OAB \) becomes:
\[ \text{Area of } OAB = \frac{1}{2} \cdot 2 |\vec{a} \times \vec{b}| = |\vec{a} \times \vec{b}| \]

- The area of triangle \( OBC \) is:
\[ \text{Area of } OBC = \frac{1}{2} |\vec{OB} \times \vec{OC}| = \frac{1}{2} |(10\vec{a} + 2\vec{b}) \times \vec{b}| \]
- Again, using the property of the cross product:
\[ (10\vec{a} + 2\vec{b}) \times \vec{b} = 10(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{b}) = 10(\vec{a} \times \vec{b}) + 0 = 10(\vec{a} \times \vec{b}) \]
- Thus, the area of triangle \( OBC \) becomes:
\[ \text{Area of } OBC = \frac{1}{2} \cdot 10 |\vec{a} \times \vec{b}| = 5 |\vec{a} \times \vec{b}| \]

4. **Combine the Areas:**
- Therefore, the total area \( P \) of quadrilateral \( OACB \) is:
\[ P = \text{Area of } OAB + \text{Area of } OBC = |\vec{a} \times \vec{b}| + 5 |\vec{a} \times \vec{b}| = 6 |\vec{a} \times \vec{b}| \]

5. **Relate Areas \( P \) and \( Q \):**
- We know that \( P = kQ \), where \( Q = |\vec{a} \times \vec{b}| \).
- Substituting the values we found:
\[ 6 |\vec{a} \times \vec{b}| = k |\vec{a} \times \vec{b}| \]

6. **Solve for \( k \):**
- Dividing both sides by \( |\vec{a} \times \vec{b}| \) (assuming it is non-zero):
\[ k = 6 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{6} \).