The number of integral values of a for which the system of linear equations ` x sin theta -2 y cos theta -az=0`, `x+2y+z=0,-x+y+z=0` may have non-trivial solutions, then

Options

A. The given system will have infinite solutions for a=2
B. The number of integer values of a is 3 for the system to have nontrivial solutions.
C. For a=1 there exists θ for which the system will have infinite solutions
D. For a=3 there exists θ for which the system will have unique solutions

Answer

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

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

### Step 1: Write the system in matrix form

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

\[ \begin{bmatrix} \sin \theta & -2 \cos \theta & -a \\ 1 & 2 & 1 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix

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

\[ D = \begin{vmatrix} \sin \theta & -2 \cos \theta & -a \\ 1 & 2 & 1 \\ -1 & 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant

We can calculate the determinant using cofactor expansion. Let's simplify the determinant:

\[ D = \sin \theta \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + 2 \cos \theta \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - a \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1 \]

\[ \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2 \]

\[ \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = (1)(1) - (2)(-1) = 1 + 2 = 3 \]

Thus, substituting back into the determinant:

\[ D = \sin \theta (1) + 2 \cos \theta (2) - a (3) = \sin \theta + 4 \cos \theta - 3a \]

### Step 4: Set the determinant to zero

For non-trivial solutions, we set the determinant to zero:

\[ \sin \theta + 4 \cos \theta - 3a = 0 \]

Rearranging gives:

\[ 3a = \sin \theta + 4 \cos \theta \]

### Step 5: Determine the range of \( a \)

The expression \( \sin \theta + 4 \cos \theta \) can be rewritten in the form \( R \sin(\theta + \phi) \), where \( R = \sqrt{1^2 + 4^2} = \sqrt{17} \).

Thus, the maximum and minimum values of \( \sin \theta + 4 \cos \theta \) are \( \sqrt{17} \) and \( -\sqrt{17} \) respectively. Therefore:

\[ -\sqrt{17} \leq \sin \theta + 4 \cos \theta \leq \sqrt{17} \]

Dividing by 3 gives:

\[ -\frac{\sqrt{17}}{3} \leq a \leq \frac{\sqrt{17}}{3} \]

### Step 6: Calculate integral values of \( a \)

Calculating \( \frac{\sqrt{17}}{3} \):

\[ \sqrt{17} \approx 4.123 \implies \frac{\sqrt{17}}{3} \approx 1.374 \]

Thus, the integral values of \( a \) that satisfy this inequality are:

\[ a = -1, 0, 1 \]

### Conclusion

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