The set of all values of $$\lambda$$ for which the system of linear equations:
$$2x_1 - 2x_2 + x_3 = \lambda x_1$$
$$2x_1 - 3x_2 + 2x_3 = \lambda x_2$$
$$-x_1 + 2x_2 = \lambda x_3$$
has a non-trivial solution,

The system of linear equations is given as:

$$2x_1 - 2x_2 + x_3 = \lambda x_1$$

$$2x_1 - 3x_2 + 2x_3 = \lambda x_2$$

$$-x_1 + 2x_2 = \lambda x_3$$

For non-trivial solutions (where not all $$x_1, x_2, x_3$$ are zero), the system must be consistent and dependent, meaning the determinant of the coefficient matrix must be zero. Rearrange each equation to bring all terms to one side:

Equation 1: $$2x_1 - 2x_2 + x_3 - \lambda x_1 = 0$$ becomes $$(2 - \lambda)x_1 - 2x_2 + x_3 = 0$$

Equation 2: $$2x_1 - 3x_2 + 2x_3 - \lambda x_2 = 0$$ becomes $$2x_1 + (-3 - \lambda)x_2 + 2x_3 = 0$$

Equation 3: $$-x_1 + 2x_2 - \lambda x_3 = 0$$ becomes $$-x_1 + 2x_2 - \lambda x_3 = 0$$

The coefficient matrix is:

$$ \begin{pmatrix} 2 - \lambda & -2 & 1 \\ 2 & -3 - \lambda & 2 \\ -1 & 2 & -\lambda \end{pmatrix} $$

Set the determinant equal to zero:

$$ \begin{vmatrix} 2 - \lambda & -2 & 1 \\ 2 & -3 - \lambda & 2 \\ -1 & 2 & -\lambda \end{vmatrix} = 0 $$

Expand along the first row:

$$ D = (2 - \lambda) \begin{vmatrix} -3 - \lambda & 2 \\ 2 & -\lambda \end{vmatrix} - (-2) \begin{vmatrix} 2 & 2 \\ -1 & -\lambda \end{vmatrix} + (1) \begin{vmatrix} 2 & -3 - \lambda \\ -1 & 2 \end{vmatrix} $$

Compute each 2x2 determinant:

First: $$\begin{vmatrix} -3 - \lambda & 2 \\ 2 & -\lambda \end{vmatrix} = (-3 - \lambda)(-\lambda) - (2)(2) = \lambda(3 + \lambda) - 4 = \lambda^2 + 3\lambda - 4$$

Second: $$\begin{vmatrix} 2 & 2 \\ -1 & -\lambda \end{vmatrix} = (2)(-\lambda) - (2)(-1) = -2\lambda + 2$$

Third: $$\begin{vmatrix} 2 & -3 - \lambda \\ -1 & 2 \end{vmatrix} = (2)(2) - (-3 - \lambda)(-1) = 4 - (3 + \lambda) = 4 - 3 - \lambda = 1 - \lambda$$

Substitute back:

$$ D = (2 - \lambda)(\lambda^2 + 3\lambda - 4) + 2(-2\lambda + 2) + (1)(1 - \lambda) $$

Expand each term:

First term: $$(2 - \lambda)(\lambda^2 + 3\lambda - 4) = 2(\lambda^2 + 3\lambda - 4) - \lambda(\lambda^2 + 3\lambda - 4) = 2\lambda^2 + 6\lambda - 8 - \lambda^3 - 3\lambda^2 + 4\lambda = -\lambda^3 - \lambda^2 + 10\lambda - 8$$

Second term: $$2(-2\lambda + 2) = -4\lambda + 4$$

Third term: $$1 - \lambda$$

Combine all:

$$ D = (-\lambda^3 - \lambda^2 + 10\lambda - 8) + (-4\lambda + 4) + (1 - \lambda) = -\lambda^3 - \lambda^2 + (10\lambda - 4\lambda - \lambda) + (-8 + 4 + 1) = -\lambda^3 - \lambda^2 + 5\lambda - 3 $$

Set $$D = 0$$:

$$ -\lambda^3 - \lambda^2 + 5\lambda - 3 = 0 $$

Multiply both sides by $$-1$$ to simplify:

$$ \lambda^3 + \lambda^2 - 5\lambda + 3 = 0 $$

Solve the cubic equation. Possible rational roots are $$\pm 1, \pm 3$$. Test $$\lambda = 1$$:

$$ (1)^3 + (1)^2 - 5(1) + 3 = 1 + 1 - 5 + 3 = 0 $$

So $$\lambda = 1$$ is a root. Factor out $$(\lambda - 1)$$ using synthetic division:

Coefficients: $$1, 1, -5, 3$$

Synthetic division with root 1:

Bring down 1, multiply by 1: $$1 \times 1 = 1$$, add to next coefficient: $$1 + 1 = 2$$, multiply by 1: $$2 \times 1 = 2$$, add to next: $$-5 + 2 = -3$$, multiply by 1: $$-3 \times 1 = -3$$, add to last: $$3 + (-3) = 0$$.

Quotient is $$\lambda^2 + 2\lambda - 3$$, so:

$$ \lambda^3 + \lambda^2 - 5\lambda + 3 = (\lambda - 1)(\lambda^2 + 2\lambda - 3) = 0 $$

Set each factor to zero:

$$\lambda - 1 = 0$$ gives $$\lambda = 1$$

$$\lambda^2 + 2\lambda - 3 = 0$$ solved by quadratic formula:

Discriminant: $$d = b^2 - 4ac = 2^2 - 4(1)(-3) = 4 + 12 = 16$$

Roots: $$\lambda = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2}$$

So $$\lambda = \frac{-2 + 4}{2} = \frac{2}{2} = 1$$ and $$\lambda = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$$

The roots are $$\lambda = 1$$ (multiplicity 2) and $$\lambda = -3$$. The distinct values are $$\lambda = -3$$ and $$\lambda = 1$$.

Verify by substitution:

For $$\lambda = -3$$:

Equations become:

$$(2 - (-3))x_1 - 2x_2 + x_3 = 0 \rightarrow 5x_1 - 2x_2 + x_3 = 0$$

$$2x_1 + (-3 - (-3))x_2 + 2x_3 = 0 \rightarrow 2x_1 + 2x_3 = 0 \rightarrow x_1 + x_3 = 0$$

$$-x_1 + 2x_2 - (-3)x_3 = 0 \rightarrow -x_1 + 2x_2 + 3x_3 = 0$$

From $$x_1 + x_3 = 0$$, set $$x_1 = -x_3$$. Substitute into first equation: $$5(-x_3) - 2x_2 + x_3 = 0 \rightarrow -5x_3 - 2x_2 + x_3 = 0 \rightarrow -2x_2 - 4x_3 = 0 \rightarrow x_2 = -2x_3$$. Substitute into third: $$-(-x_3) + 2(-2x_3) + 3x_3 = x_3 - 4x_3 + 3x_3 = 0$$, which holds. Non-trivial solution exists (e.g., $$x_3 = 1, x_1 = -1, x_2 = -2$$).

For $$\lambda = 1$$:

Equations become:

$$(2 - 1)x_1 - 2x_2 + x_3 = 0 \rightarrow x_1 - 2x_2 + x_3 = 0$$

$$2x_1 + (-3 - 1)x_2 + 2x_3 = 0 \rightarrow 2x_1 - 4x_2 + 2x_3 = 0 \rightarrow x_1 - 2x_2 + x_3 = 0$$ (same as first)

$$-x_1 + 2x_2 - (1)x_3 = 0 \rightarrow -x_1 + 2x_2 - x_3 = 0$$, which is equivalent to $$x_1 - 2x_2 + x_3 = 0$$ (multiply by $$-1$$). So only one independent equation: $$x_1 - 2x_2 + x_3 = 0$$. Non-trivial solutions exist (e.g., $$x_2 = 1, x_3 = 0, x_1 = 2$$).

The set of $$\lambda$$ values is $$\{-3, 1\}$$, which has two distinct elements.

Options:

A. Contains more than two elements.

B. Is an empty set.

C. Is a singleton.

D. Contains two elements.

Hence, the correct answer is Option D.