If A,B and C are the angles of a triangle and
`|{:(1,1,1),(1+sinA,1+sinB,1+sinC),(sinA+sin^(2)A,sinB+sin^(2)B,sinCsin^(2)C):}|` =0
then `Delta` ABC must be .

Options

A. right angled isosceles
B. isosceles
C. equilateral
D. scalene

Answer

To solve the given problem, we need to analyze the determinant and apply properties of determinants to find the relationship between the angles A, B, and C of the triangle.

### Step-by-Step Solution:

1. **Write the Determinant**:
We start with the determinant:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \sin A & 1 + \sin B & 1 + \sin C \\ \sin A + \sin^2 A & \sin B + \sin^2 B & \sin C + \sin^2 C \end{vmatrix} \]
Given that \( D = 0 \).

2. **Apply Column Operations**:
We can perform column operations to simplify the determinant. Let's subtract the first column from the second and third columns:
\[ D = \begin{vmatrix} 1 & 0 & 0 \\ 1 + \sin A & \sin B - \sin A & \sin C - \sin A \\ \sin A + \sin^2 A & \sin^2 B - \sin^2 A & \sin^2 C - \sin^2 A \end{vmatrix} \]

3. **Simplify the Determinant**:
The determinant now looks like:
\[ D = \begin{vmatrix} 1 & 0 & 0 \\ 1 + \sin A & \sin B - \sin A & \sin C - \sin A \\ \sin A + \sin^2 A & \sin^2 B - \sin^2 A & \sin^2 C - \sin^2 A \end{vmatrix} \]
Since the first column has been simplified, we can expand the determinant along the first column.

4. **Factor Out Common Terms**:
We can factor out \( \sin B - \sin A \) from the second column and \( \sin C - \sin A \) from the third column:
\[ D = (\sin B - \sin A)(\sin C - \sin A) \cdot \begin{vmatrix} 1 & 1 \\ 1 + \sin A & 1 + \sin A \end{vmatrix} \]

5. **Evaluate the Determinant**:
The remaining determinant simplifies to:
\[ D = (\sin B - \sin A)(\sin C - \sin A) \cdot 0 = 0 \]
This indicates that at least one of the factors must be zero.

6. **Conclusion**:
Since \( D = 0 \), we have:
\[ \sin A = \sin B \quad \text{or} \quad \sin B = \sin C \quad \text{or} \quad \sin A = \sin C \]
This implies that at least two angles of the triangle are equal, which means triangle ABC is isosceles.

### Final Answer:
The triangle ABC must be **isosceles**.