Let `lambda` and `alpha` be real. Then the numbers of intergral values `lambda` for which the system of linear equations
`lambdax +(sin alpha) y+ (cos alpha) z=0`
`x + (cos alpha) y+ (sin alpha) z=0`
`-x+(sin alpha) y -(cos alpha) z=0` has non-trivial solutions is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to determine the number of integral values of \(\lambda\) for which the given system of linear equations has non-trivial solutions. The equations are:

1. \(\lambda x + (\sin \alpha) y + (\cos \alpha) z = 0\)
2. \(x + (\cos \alpha) y + (\sin \alpha) z = 0\)
3. \(-x + (\sin \alpha) y - (\cos \alpha) z = 0\)

### Step-by-Step Solution

**Step 1: Write the system in matrix form.**

We can express the system of equations in matrix form as follows:

\[ \begin{bmatrix} \lambda & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

**Step 2: Set up the determinant condition for non-trivial solutions.**

For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero:

\[ \text{Det} = \begin{vmatrix} \lambda & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \end{vmatrix} = 0 \]

**Step 3: Calculate the determinant.**

We can calculate the determinant using the cofactor expansion along the first row:

\[ \text{Det} = \lambda \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} - \sin \alpha \begin{vmatrix} 1 & \sin \alpha \\ -1 & -\cos \alpha \end{vmatrix} + \cos \alpha \begin{vmatrix} 1 & \cos \alpha \\ -1 & \sin \alpha \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -\cos^2 \alpha - \sin^2 \alpha = -1\)

2. \(\begin{vmatrix} 1 & \sin \alpha \\ -1 & -\cos \alpha \end{vmatrix} = -\cos \alpha + \sin \alpha = \sin \alpha - \cos \alpha\)

3. \(\begin{vmatrix} 1 & \cos \alpha \\ -1 & \sin \alpha \end{vmatrix} = \sin \alpha + \cos \alpha\)

Putting it all together:

\[ \text{Det} = \lambda (-1) - \sin \alpha (\sin \alpha - \cos \alpha) + \cos \alpha (\sin \alpha + \cos \alpha) \]

**Step 4: Simplify the determinant expression.**

\[ \text{Det} = -\lambda - \sin^2 \alpha + \sin \alpha \cos \alpha + \cos \alpha \sin \alpha + \cos^2 \alpha \]
\[ = -\lambda - \sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha \]
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\[ = -\lambda + 1 + 2 \sin \alpha \cos \alpha \]
\[ = -\lambda + 1 + \sin 2\alpha \]

**Step 5: Set the determinant equal to zero.**

Setting the determinant to zero gives:

\[ -\lambda + 1 + \sin 2\alpha = 0 \]
\[ \lambda = 1 + \sin 2\alpha \]

**Step 6: Determine the range of \(\lambda\).**

The range of \(\sin 2\alpha\) is \([-1, 1]\), thus:

\[ \lambda = 1 + \sin 2\alpha \implies \lambda \in [0, 2] \]

**Step 7: Find the integral values of \(\lambda\).**

The integral values of \(\lambda\) in the interval \([0, 2]\) are:

- \(0\)
- \(1\)
- \(2\)

Thus, there are **3 integral values** of \(\lambda\).

### Final Answer

The number of integral values of \(\lambda\) for which the system has non-trivial solutions is **3**.