Consider the system of equations : `x sintheta-2ycostheta-az=0`, `x+2y+z=0`, `-x+y+z=0`, `theta in R`

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 `theta` for which the system will have infinite solutions
D. For `a=3` there exists `theta` for which the system will have unique solutions

Explanation

To solve the given system of equations, we will first rewrite the equations in a standard form and then represent them in a determinant format.

### Step 1: Write the equations in matrix form
The given equations are:
1. \( x \sin \theta - 2y \cos \theta - az = 0 \)
2. \( x + 2y + z = 0 \)
3. \( -x + y + z = 0 \)

We can express these equations in the form of a matrix \( A \) and a vector \( \mathbf{X} \):
\[ \begin{bmatrix} \sin \theta & -2 \cos \theta & -a \\ 1 & 2 & 1 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set up the determinant
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

Calculating the determinant:
\[ \text{det}(A) = \begin{vmatrix} \sin \theta & -2 \cos \theta & -a \\ 1 & 2 & 1 \\ -1 & 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - eg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

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

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 1 = 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot (-1) = 2 \)
3. \( \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot (-1) = 3 \)

Putting it all together:
\[ \text{det}(A) = \sin \theta \cdot 1 + 2 \cos \theta \cdot 2 - a \cdot 3 \]
\[ = \sin \theta + 4 \cos \theta - 3a \]

### Step 4: Set the determinant to zero
Setting the determinant to zero for non-trivial solutions:
\[ \sin \theta + 4 \cos \theta - 3a = 0 \]
Rearranging gives:
\[ 3a = \sin \theta + 4 \cos \theta \]
\[ a = \frac{\sin \theta + 4 \cos \theta}{3} \]

### Step 5: Analyze the range of \( a \)
To find the range of \( a \), we need to find the maximum and minimum values of \( \sin \theta + 4 \cos \theta \).

Using the method of Lagrange multipliers or the Cauchy-Schwarz inequality, we can find that:
\[ \sqrt{1^2 + 4^2} = \sqrt{17} \]
Thus, the maximum value of \( \sin \theta + 4 \cos \theta \) is \( \sqrt{17} \) and the minimum value is \( -\sqrt{17} \).

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

Dividing by 3 gives:
\[ -\frac{\sqrt{17}}{3} \leq a \leq \frac{\sqrt{17}}{3} \]

### Conclusion
The values of \( a \) that satisfy the system of equations are within the range:
\[ a \in \left[-\frac{\sqrt{17}}{3}, \frac{\sqrt{17}}{3}\right] \]