`a,b,c,in N and d = |{:(a,b,c),(c,a,b),(b,c,a):}|` , then the least positive value of d is

Options

A. 4
B. 6
C. 3
D. 8

Answer

To find the least positive value of \( d \), where \( d = | \{ (a,b,c), (c,a,b), (b,c,a) \} | \) and \( a, b, c \in \mathbb{N} \), we will evaluate the determinant of the matrix formed by the permutations of \( a, b, c \).

1. **Set up the determinant**:
\[ d = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} \]

2. **Calculate the determinant**:
Using the determinant formula for a 3x3 matrix:
\[ d = a \begin{vmatrix} a & b \\ c & a \end{vmatrix} - b \begin{vmatrix} c & b \\ b & a \end{vmatrix} + c \begin{vmatrix} c & a \\ b & c \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
- \( \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \)
- \( \begin{vmatrix} c & b \\ b & a \end{vmatrix} = ca - b^2 \)
- \( \begin{vmatrix} c & a \\ b & c \end{vmatrix} = c^2 - ab \)

Substitute these back into the determinant:
\[ d = a(a^2 - bc) - b(ca - b^2) + c(c^2 - ab) \]
Expanding this gives:
\[ d = a^3 - abc - bca + b^3 + c^3 - abc \]
Combining like terms results in:
\[ d = a^3 + b^3 + c^3 - 3abc \]

3. **Finding the least positive value of \( d \)**:
We need to find the least positive value of \( d \) when \( a, b, c \) are natural numbers.

- Start with the smallest values for \( a, b, c \). Let’s try \( a = 1, b = 1, c = 1 \):
\[ d = 1^3 + 1^3 + 1^3 - 3 \cdot 1 \cdot 1 \cdot 1 = 3 - 3 = 0 \]
(This is not positive.)

- Next, try \( a = 1, b = 1, c = 2 \):
\[ d = 1^3 + 1^3 + 2^3 - 3 \cdot 1 \cdot 1 \cdot 2 = 1 + 1 + 8 - 6 = 4 \]
(This is positive.)

- Check if we can get a smaller positive value by trying other combinations:
- \( a = 1, b = 2, c = 2 \):
\[ d = 1^3 + 2^3 + 2^3 - 3 \cdot 1 \cdot 2 \cdot 2 = 1 + 8 + 8 - 12 = 5 \]
- \( a = 2, b = 2, c = 2 \):
\[ d = 2^3 + 2^3 + 2^3 - 3 \cdot 2 \cdot 2 \cdot 2 = 8 + 8 + 8 - 24 = 0 \]

The smallest positive value we found is \( d = 4 \).

Thus, the least positive value of \( d \) is \( \boxed{4} \).