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.
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