以原点 $O$ 为参考，点 $A$、$B$ 和 $C$ 满足 $\overrightarrow{OA} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$、$\overrightarrow{OB} = \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}$ 和 $\overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$。点 $P$ 在 $AB$ 上，介于 $A$ 和 $B$ 之间，使得 $AP : PB = \lambda : 1-\lambda$，其中 $0 < \lambda < 1$。点 $Q$ 在 $BC$ 上，介于 $B$ 和 $C$ 之间，使得 $BQ : QC = \lambda : 1-\lambda$。

1. Find the position vector of point $P$ using the Ratio Theorem.

Using the Ratio Theorem since $AP : PB = \lambda : 1-\lambda$,
$\overrightarrow{OP} = (1-\lambda)\overrightarrow{OA} + \lambda\overrightarrow{OB}$
$\overrightarrow{OP} = (1-\lambda)\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 - 2\lambda \\ 1 - \lambda + \lambda \\ -1 + \lambda - 2\lambda \end{pmatrix} = \begin{pmatrix} 2 - 2\lambda \\ 1 \\ -1 - \lambda \end{pmatrix}$

2. Find the position vector of point $Q$ using the Ratio Theorem.

Similarly, since $BQ : QC = \lambda : 1-\lambda$,
$\overrightarrow{OQ} = (1-\lambda)\overrightarrow{OB} + \lambda\overrightarrow{OC}$
$\overrightarrow{OQ} = (1-\lambda)\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} = \begin{pmatrix} \lambda \\ 1 - \lambda + 3\lambda \\ -2 + 2\lambda \end{pmatrix} = \begin{pmatrix} \lambda \\ 1 + 2\lambda \\ -2 + 2\lambda \end{pmatrix}$

3. Calculate the displacement vector $\overrightarrow{PQ}$.

$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$
$\overrightarrow{PQ} = \begin{pmatrix} \lambda \\ 1 + 2\lambda \\ -2 + 2\lambda \end{pmatrix} - \begin{pmatrix} 2 - 2\lambda \\ 1 \\ -1 - \lambda \end{pmatrix}$
$\overrightarrow{PQ} = \begin{pmatrix} \lambda - (2 - 2\lambda) \\ 1 + 2\lambda - 1 \\ -2 + 2\lambda - (-1 - \lambda) \end{pmatrix} = \begin{pmatrix} 3\lambda - 2 \\ 2\lambda \\ 3\lambda - 1 \end{pmatrix}$

1. Determine the displacement vector $\overrightarrow{PX}$ using the position vector $\overrightarrow{OP}$ found in part (a)

From part (a), the position vector of $P$ is $\overrightarrow{OP} = \begin{pmatrix} 2-2\lambda \\ 1 \\ -1-\lambda \end{pmatrix}$ and $\overrightarrow{PQ} = \begin{pmatrix} 3\lambda - 2 \\ 2\lambda \\ 3\lambda - 1 \end{pmatrix}$.

The displacement vector $\overrightarrow{PX}$ is given by:
$\overrightarrow{PX} = \overrightarrow{OX} - \overrightarrow{OP} = \begin{pmatrix} 2 \\ -1 \\ -\frac{5}{2} \end{pmatrix} - \begin{pmatrix} 2 - 2\lambda \\ 1 \\ -1 - \lambda \end{pmatrix} = \begin{pmatrix} 2\lambda \\ -2 \\ \lambda - \frac{3}{2} \end{pmatrix}$

2. Use the condition for collinearity to set up equations involving $\lambda$

Since points $P$, $Q$, and $X$ are collinear, $\overrightarrow{PX}$ is a scalar multiple of $\overrightarrow{PQ}$. Therefore, there exists a scalar $k$ such that:
$\overrightarrow{PX} = k\overrightarrow{PQ}$
$\begin{pmatrix} 2\lambda \\ -2 \\ \lambda - \frac{3}{2} \end{pmatrix} = k \begin{pmatrix} 3\lambda - 2 \\ 2\lambda \\ 3\lambda - 1 \end{pmatrix}$
Equating the $y$-components:
$-2 = k(2\lambda) \implies k = -\frac{1}{\lambda}$

3. Solve the equations to find the value of $\lambda$, taking the domain into account

Equating the $x$-components and substituting $k = -\frac{1}{\lambda}$:
$2\lambda = \left(-\frac{1}{\lambda}\right)(3\lambda - 2)$
$2\lambda^2 = -3\lambda + 2$
$2\lambda^2 + 3\lambda - 2 = 0$
$(2\lambda - 1)(\lambda + 2) = 0$
$\lambda = \frac{1}{2} \quad \text{or} \quad \lambda = -2$
Since it is given that $0 < \lambda < 1$, we reject $\lambda = -2$.
$\lambda = \frac{1}{2}$