Let `|veca|=1, |vecb|=1 and |veca+vecb|=sqrt(3)`. If `vec c` be a vector such that `vec c=veca+2vecb-3(veca xx vecb)` and `p=|(veca xx vecb) xx vec c|`, then find `[p^(2)]`. (where [ ] represents greatest integer function).

Answer

To solve the problem, we will follow a systematic approach step by step.

### Step 1: Analyze the given information
We have the following:
- \( |\vec{a}| = 1 \)
- \( |\vec{b}| = 1 \)
- \( |\vec{a} + \vec{b}| = \sqrt{3} \)

### Step 2: Use the properties of vectors
From the property of magnitudes, we know:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]
Substituting the known values:
\[ (\sqrt{3})^2 = 1^2 + 1^2 + 2 \vec{a} \cdot \vec{b} \]
This simplifies to:
\[ 3 = 1 + 1 + 2 \vec{a} \cdot \vec{b} \]
\[ 3 = 2 + 2 \vec{a} \cdot \vec{b} \]
\[ 2 \vec{a} \cdot \vec{b} = 1 \implies \vec{a} \cdot \vec{b} = \frac{1}{2} \]

### Step 3: Define vector \( \vec{c} \)
We are given:
\[ \vec{c} = \vec{a} + 2\vec{b} - 3(\vec{a} \times \vec{b}) \]

### Step 4: Find \( p = |\vec{a} \times \vec{b} \times \vec{c}| \)
Using the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} \]
Let \( \vec{x} = \vec{a} \times \vec{b} \), \( \vec{y} = \vec{c} \), and \( \vec{z} = \vec{a} \times \vec{b} \):
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})(\vec{b}) - (\vec{a} \cdot \vec{b})(\vec{c}) \]

### Step 5: Calculate \( \vec{a} \cdot \vec{c} \)
Calculating \( \vec{a} \cdot \vec{c} \):
\[ \vec{a} \cdot \vec{c} = \vec{a} \cdot \left( \vec{a} + 2\vec{b} - 3(\vec{a} \times \vec{b}) \right) \]
\[ = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} - 3 \vec{a} \cdot (\vec{a} \times \vec{b}) \]
Since \( \vec{a} \cdot (\vec{a} \times \vec{b}) = 0 \):
\[ = 1 + 2 \cdot \frac{1}{2} = 1 + 1 = 2 \]

### Step 6: Calculate \( \vec{b} \cdot \vec{c} \)
Calculating \( \vec{b} \cdot \vec{c} \):
\[ \vec{b} \cdot \vec{c} = \vec{b} \cdot \left( \vec{a} + 2\vec{b} - 3(\vec{a} \times \vec{b}) \right) \]
\[ = \vec{b} \cdot \vec{a} + 2 \vec{b} \cdot \vec{b} - 3 \vec{b} \cdot (\vec{a} \times \vec{b}) \]
Again, \( \vec{b} \cdot (\vec{a} \times \vec{b}) = 0 \):
\[ = \frac{1}{2} + 2 \cdot 1 = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \]

### Step 7: Calculate \( p \)
Now substituting back:
\[ p = |\vec{a} \cdot \vec{c} \vec{b} - \vec{a} \cdot \vec{b} \vec{c}| \]
\[ = |2 \vec{b} - \frac{1}{2} \vec{c}| \]
Substituting \( \vec{c} \):
\[ = |2 \vec{b} - \frac{1}{2} (\vec{a} + 2\vec{b} - 3(\vec{a} \times \vec{b}) )| \]
Calculating this gives us:
\[ = |2 \vec{b} - \frac{1}{2} \vec{a} - \vec{b} + \frac{3}{2}(\vec{a} \times \vec{b})| \]
\[ = |\frac{3}{2} \vec{b} - \frac{1}{2} \vec{a} + \frac{3}{2}(\vec{a} \times \vec{b})| \]

### Step 8: Find \( p^2 \)
Calculating \( p^2 \):
\[ p^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 + \left( \frac{3}{2} \right)^2 \]
\[ = \frac{9}{4} + \frac{1}{4} + \frac{9}{4} = \frac{19}{4} \]

### Step 9: Apply the greatest integer function
The greatest integer function \( [p^2] = [\frac{19}{4}] = [4.75] = 4 \).

### Final Answer
Thus, the answer is:
\[ \boxed{4} \]