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, we will follow these steps:

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

### Step 2: Analyze the dot product condition
The condition given is \( (\vec{c} - \vec{a}) \cdot (\vec{b} - \vec{c}) = 0 \). This implies that the vectors \( \vec{c} - \vec{a} \) and \( \vec{b} - \vec{c} \) are perpendicular. Therefore, triangle \( ABC \) is a right triangle with the right angle at vertex \( C \).

### Step 3: Apply the Pythagorean theorem
Since \( C \) is the right angle, we can apply the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]
Substituting the lengths:
\[ (k + 2)^2 = (k + 1)^2 + k^2 \]
Expanding each term:
\[ k^2 + 4k + 4 = (k^2 + 2k + 1) + k^2 \]
Simplifying the right-hand side:
\[ k^2 + 4k + 4 = 2k^2 + 2k + 1 \]
Rearranging gives:
\[ 0 = 2k^2 + 2k + 1 - k^2 - 4k - 4 \]
\[ 0 = k^2 - 2k - 3 \]
Factoring:
\[ 0 = (k - 3)(k + 1) \]
Thus, \( k = 3 \) (since \( k \) must be positive).

### Step 4: Determine the lengths of the sides
Now substituting \( k = 3 \):
- \( BC = 3 \)
- \( CA = 4 \)
- \( AB = 5 \)

### Step 5: Calculate the area of triangle \( ABC \)
The area \( A \) of triangle \( ABC \) can be calculated using the formula for the area of a right triangle:
\[ A = \frac{1}{2} \times BC \times CA = \frac{1}{2} \times 3 \times 4 = 6 \]

### Step 6: Calculate the value of \( |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \)
The magnitude of the vector sum of the cross products can be interpreted as six times the area of the triangle:
\[ |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 2 \times \text{Area of } \triangle ABC \]
Thus:
\[ |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 2 \times 6 = 12 \]

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