If in a `Delta ABC, sin ^(3) A + sin ^(3) B+ sin ^(3) C`
`=3 sin A .Sin B. sin C,` then find the valueof determinant
`|{:(a,b,c),(b,c,a),(c,a,b):}|.`

Answer

To solve the problem, we need to find the value of the determinant given the condition involving the sine of the angles in triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Given Condition:**
We are given that:
\[ \sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C \]
This condition is satisfied when \( A + B + C = 180^\circ \) (the angles of a triangle).

2. **Using the Sine Rule:**
According to the sine rule in a triangle:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
Thus, we can express the sides \( a, b, c \) in terms of \( k \):
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

3. **Substituting into the Given Condition:**
Substitute \( a, b, c \) into the given condition:
\[ \left(\frac{a}{k}\right)^3 + \left(\frac{b}{k}\right)^3 + \left(\frac{c}{k}\right)^3 = 3 \left(\frac{a}{k}\right) \left(\frac{b}{k}\right) \left(\frac{c}{k}\right) \]
This simplifies to:
\[ \frac{a^3 + b^3 + c^3}{k^3} = \frac{3abc}{k^3} \]
Therefore, we have:
\[ a^3 + b^3 + c^3 = 3abc \]

4. **Using the Identity:**
The identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \) implies that:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
This means either \( a + b + c = 0 \) or \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \). Since \( a, b, c \) are lengths of sides of a triangle, \( a + b + c \neq 0 \). Thus:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 0 \]

5. **Finding the Value of the Determinant:**
The determinant we need to evaluate is:
\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Expanding this determinant using the first row:
\[ D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating each of the 2x2 determinants:
\[ D = a(cb - a^2) - b(bb - ac) + c(ab - c^2) \]
Simplifying this gives:
\[ D = acb - a^3 - b^3 + abc - c^3 \]
Combining terms, we find:
\[ D = 3abc - (a^3 + b^3 + c^3) \]
Since \( a^3 + b^3 + c^3 = 3abc \), we have:
\[ D = 3abc - 3abc = 0 \]

### Final Answer:
Thus, the value of the determinant is:
\[ \boxed{0} \]