In a `DeltaABC`, the sides BC, CA and AB are consecutive positive integers in increasing order. Let `veca, vecb and vecc` are position vectors of the vertices A, B and C respectively. If `(vecc-veca).(vecb-vecc)=0`, then the value of `|veca xx vecb+vecbxx vecc+vecc xxveca|` is equal to

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript while providing a clear mathematical approach.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a triangle \( ABC \) with sides \( BC \), \( CA \), and \( AB \) being consecutive positive integers in increasing order. Let's denote the lengths of these sides as follows:
- \( BC = k \)
- \( CA = k + 1 \)
- \( AB = k + 2 \)

2. **Position Vectors**:
Let \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) be the position vectors of points \( A \), \( B \), and \( C \) respectively.

3. **Given Condition**:
We have the condition:
\[ (\vec{c} - \vec{a}) \cdot (\vec{b} - \vec{c}) = 0 \]
This implies that the vectors \( \vec{c} - \vec{a} \) and \( \vec{b} - \vec{c} \) are perpendicular, indicating that triangle \( ABC \) is a right triangle.

4. **Applying Pythagorean Theorem**:
Since \( ABC \) is a right triangle, we can apply the Pythagorean theorem:
\[ AB^2 = BC^2 + CA^2 \]
Substituting the lengths:
\[ (k + 2)^2 = k^2 + (k + 1)^2 \]

5. **Expanding the Equation**:
Expanding both sides:
\[ k^2 + 4k + 4 = k^2 + (k^2 + 2k + 1) \]
Simplifying the right side:
\[ k^2 + 4k + 4 = 2k^2 + 2k + 1 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 0 = 2k^2 + 2k + 1 - k^2 - 4k - 4 \]
Simplifying this leads to:
\[ 0 = k^2 - 2k - 3 \]

7. **Factoring the Quadratic**:
Factoring the quadratic equation:
\[ (k - 3)(k + 1) = 0 \]
This gives us two potential solutions for \( k \):
\[ k = 3 \quad \text{or} \quad k = -1 \]
Since \( k \) must be positive, we take \( k = 3 \).

8. **Finding the Side Lengths**:
Now substituting \( k = 3 \):
- \( BC = 3 \)
- \( CA = 4 \)
- \( AB = 5 \)

9. **Calculating the Area**:
The area \( A \) of triangle \( ABC \) can also be calculated using the formula for the area of a right triangle:
\[ A = \frac{1}{2} \times BC \times CA = \frac{1}{2} \times 3 \times 4 = 6 \]

10. **Relating Area to Cross Products**:
The area can also be expressed in terms of the cross products of the position vectors:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \]
Therefore, we have:
\[ |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 12 \]

### Final Answer:
The value of \( |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \) is \( \boxed{12} \).