If A ,B and C are the angles of an equilateral triangle, then the value of
`|{:(1,1,1),(1+sinA,1+sinB,1+sinC),(sinA+sin^2A,sinB+sin^2B,sinC+sin^2C):}|` is ...........

Answer

To solve the given determinant problem, we will follow a systematic approach.

Given the determinant:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \sin A & 1 + \sin B & 1 + \sin C \\ \sin A + \sin^2 A & \sin B + \sin^2 B & \sin C + \sin^2 C \end{vmatrix} \]

### Step 1: Recognize the angles of the equilateral triangle
Since A, B, and C are the angles of an equilateral triangle, we have:

\[ A = B = C = \frac{\pi}{3} \]

### Step 2: Substitute the values of sin A, sin B, and sin C
Using the fact that \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), we can substitute:

\[ \sin A = \sin B = \sin C = \frac{\sqrt{3}}{2} \]

### Step 3: Substitute the values into the determinant
Now we can substitute these values into the determinant:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \frac{\sqrt{3}}{2} & 1 + \frac{\sqrt{3}}{2} & 1 + \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 & \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 & \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 \end{vmatrix} \]

Calculating \(\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\), we have:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \frac{\sqrt{3}}{2} & 1 + \frac{\sqrt{3}}{2} & 1 + \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} + \frac{3}{4} & \frac{\sqrt{3}}{2} + \frac{3}{4} & \frac{\sqrt{3}}{2} + \frac{3}{4} \end{vmatrix} \]

### Step 4: Simplify the determinant
Notice that all rows are proportional. The second and third rows are identical:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \frac{\sqrt{3}}{2} & 1 + \frac{\sqrt{3}}{2} & 1 + \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} + \frac{3}{4} & \frac{\sqrt{3}}{2} + \frac{3}{4} & \frac{\sqrt{3}}{2} + \frac{3}{4} \end{vmatrix} \]

### Step 5: Conclude the determinant value
Since the second and third rows are identical, the determinant evaluates to zero:

\[ D = 0 \]

Thus, the value of the determinant is:

\[ \boxed{0} \]