If the maximum and minimum values of the determinant
`|(1 + sin^(2)x,cos^(2) x,sin 2x),(sin^(2) x,1 + cos^(2) x,sin 2x),(sin^(2) x,cos^(2) x,1 + sin 2x)|` are `alpha and beta`, then

Options

A. `alpha+beta^(99)=4`
B. `alpha^(3)-beta^(17)=26`
C. `(alpha^(2n)-beta^(2n))` is always an even integer for `n in N`
D. a triangle can be drawn having it's sides as `alpha,betaand alpha-beta`

Answer

To solve the given problem, we need to evaluate the determinant and find its maximum and minimum values. Let's break down the solution step by step.

### Step 1: Write the Determinant
We start with the determinant:

\[ D = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

### Step 2: Simplify the Determinant
We can perform column operations to simplify the determinant. Let's replace the first column \(C_1\) with \(C_1 + C_2\):

\[ C_1 \rightarrow C_1 + C_2 \]

This gives us:

\[ D = \begin{vmatrix} 2 + \sin^2 x & \cos^2 x & \sin 2x \\ 1 + \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

### Step 3: Further Simplification
Now, we can perform a row operation \(R_2 \rightarrow R_2 - R_1\):

\[ R_2 \rightarrow R_2 - R_1 \]

This results in:

\[ D = \begin{vmatrix} 2 + \sin^2 x & \cos^2 x & \sin 2x \\ -1 + \sin^2 x & 1 + \cos^2 x - \cos^2 x & 0 \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we can calculate the determinant using the second row:

\[ D = (-1 + \sin^2 x) \cdot \begin{vmatrix} \cos^2 x & \sin 2x \\ \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} \cos^2 x & \sin 2x \\ \cos^2 x & 1 + \sin 2x \end{vmatrix} = \cos^2 x (1 + \sin 2x) - \sin 2x \cdot \cos^2 x = \cos^2 x \]

Thus, we have:

\[ D = (-1 + \sin^2 x) \cdot \cos^2 x \]

### Step 5: Find Maximum and Minimum Values
The expression for the determinant simplifies to:

\[ D = \cos^2 x \cdot (\sin^2 x - 1) = -\cos^2 x \cdot (1 - \sin^2 x) = -\cos^2 x \cdot \cos^2 x = -\cos^4 x \]

The maximum value of \(-\cos^4 x\) occurs when \(\cos^4 x\) is minimized, which is at \(0\) (when \(\cos x = 0\)), giving us a maximum value of \(0\). The minimum value occurs when \(\cos^4 x = 1\) (when \(\cos x = \pm 1\)), giving us a minimum value of \(-1\).

Thus, we have:

\[ \alpha = 0 \quad \text{and} \quad \beta = -1 \]

### Step 6: Final Relations
Now we need to check the given relations:

1. \(\alpha + \beta = 0 - 1 = -1\)
2. \(\alpha^3 - \beta^{17} = 0^3 - (-1)^{17} = 0 + 1 = 1\)
3. \(\alpha^{2n} - \beta^{2n} = 0 - 1 = -1\) (which is not always an even integer).

### Conclusion
The values of \(\alpha\) and \(\beta\) are \(0\) and \(-1\) respectively. The relationships provided in the question can be evaluated based on these values.