The set of all values of $$\lambda$$ for which the system of linear equations $$x - 2y - 2z = \lambda x$$, $$x + 2y + z = \lambda y$$, $$-x - y = \lambda z$$ has a non-trivial solution:

First, we rewrite each given equation so that all the variables appear on the left-hand side and the constant term on the right becomes zero. This makes the system homogeneous.

$$\begin{aligned} x-2y-2z &= \lambda x \;&\Longrightarrow&\;(1-\lambda)x-2y-2z=0,\\[4pt] x+2y+z &= \lambda y \;&\Longrightarrow&\;x+(2-\lambda)y+z=0,\\[4pt] -x-y &= \lambda z \;&\Longrightarrow&\;-x-y-\lambda z=0. \end{aligned}$$

We now express the system in matrix form $$A\mathbf{x}=\mathbf{0}$$ where $$\mathbf{x}=\begin{bmatrix}x\\y\\z\end{bmatrix}$$ and the coefficient matrix $$A$$ depends on $$\lambda$$:

$$ A= \begin{bmatrix} 1-\lambda & -2 & -2\\ 1 & 2-\lambda & 1\\ -1 & -1 & -\lambda \end{bmatrix}. $$

For a homogeneous linear system to have a non-trivial solution, a necessary and sufficient condition is that the determinant of its coefficient matrix be zero. We therefore set $$\det A=0$$.

Using the standard $$3\times3$$ determinant expansion, we have

$$ \det A= \begin{vmatrix} 1-\lambda & -2 & -2\\ 1 & 2-\lambda & 1\\ -1 & -1 & -\lambda \end{vmatrix}. $$

Expanding along the first row:

$$ \begin{aligned} \det A &= (1-\lambda) \Big[(2-\lambda)(-\lambda)-1(-1)\Big] -(-2)\Big[1(-\lambda)-1(-1)\Big] +(-2)\Big[1(-1)-(2-\lambda)(-1)\Big]. \end{aligned} $$

We simplify each bracketed term carefully.

$$ \begin{aligned} (2-\lambda)(-\lambda)-1(-1) &= -\lambda(2-\lambda)+1 = -2\lambda+\lambda^{2}+1 = \lambda^{2}-2\lambda+1 = (\,\lambda-1\,)^{2},\\[6pt] 1(-\lambda)-1(-1) &= -\lambda+1 = 1-\lambda,\\[6pt] 1(-1)-(2-\lambda)(-1) &= -1+\,(2-\lambda) = 1-\lambda. \end{aligned} $$

Substituting these back:

$$ \begin{aligned} \det A &= (1-\lambda)\bigl(\lambda-1\bigr)^{2} +2\,(1-\lambda) -2\,(1-\lambda). \end{aligned} $$

The last two terms cancel each other, leaving

$$ \det A = (1-\lambda)\bigl(\lambda-1\bigr)^{2} = (1-\lambda)\,(1-\lambda)^{2} = (1-\lambda)^{3}. $$

For a non-trivial solution we require $$\det A=0$$, so

$$ (1-\lambda)^{3}=0 \;\Longrightarrow\; 1-\lambda=0 \;\Longrightarrow\; \lambda=1. $$

Thus there is exactly one value of $$\lambda$$ that yields a non-trivial solution. The set of all such $$\lambda$$ is therefore a singleton.

Hence, the correct answer is Option C.