如上图所示的平行四边形 $OACB$ 中，$\overrightarrow{OA} = \mathbf{a}$ 且 $\overrightarrow{OB} = \mathbf{b}$。已知 $M$ 是 $AC$ 的中点，且 $ON : NB = 2 : 1$。点 $P$ 是直线 $OM$ 与 $AN$ 的交点。

1. Find an expression for $\overrightarrow{OP}$ in terms of $\lambda$, $\mathbf{a}$, and $\mathbf{b}$.

Given that $M$ is the midpoint of $AC$, and $OACB$ is a parallelogram:
$\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}\overrightarrow{OB} = \frac{1}{2}\mathbf{b}$
$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \frac{1}{2}\mathbf{b}$
Since $\overrightarrow{OP} = \lambda \overrightarrow{OM}$, we have:
$\overrightarrow{OP} = \lambda \left( \mathbf{a} + \frac{1}{2}\mathbf{b} \right) = \lambda \mathbf{a} + \frac{\lambda}{2}\mathbf{b}$

2. Find a second expression for $\overrightarrow{OP}$ in terms of $\mu$, $\mathbf{a}$, and $\mathbf{b}$.

Given that $ON : NB = 2 : 1$, we have $\overrightarrow{ON} = \frac{2}{3}\overrightarrow{OB} = \frac{2}{3}\mathbf{b}$.
$\overrightarrow{AN} = \overrightarrow{AO} + \overrightarrow{ON} = -\mathbf{a} + \frac{2}{3}\mathbf{b}$
Since $\overrightarrow{AP} = \mu \overrightarrow{AN}$, we have:
$\overrightarrow{AP} = \mu \left( -\mathbf{a} + \frac{2}{3}\mathbf{b} \right)$
$\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \mu \left( -\mathbf{a} + \frac{2}{3}\mathbf{b} \right) = (1 - \mu)\mathbf{a} + \frac{2\mu}{3}\mathbf{b}$

3. Form simultaneous equations by equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$.

Since vectors $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, we can equate their respective coefficients from the two expressions for $\overrightarrow{OP}$:
For $\mathbf{a}$: $\lambda = 1 - \mu$ (Equation 1)
For $\mathbf{b}$: $\frac{\lambda}{2} = \frac{2\mu}{3}$ (Equation 2)

4. Solve the simultaneous equations to find the values of $\lambda$ and $\mu$.

From Equation 2, we have $\lambda = \frac{4\mu}{3}$.
Substituting this into Equation 1 gives:
$\frac{4\mu}{3} = 1 - \mu$
$\frac{4\mu}{3} + \mu = 1$
$\frac{7\mu}{3} = 1 \implies \mu = \frac{3}{7}$
Substituting $\mu$ back into the expression for $\lambda$:
$\lambda = \frac{4}{3}\left(\frac{3}{7}\right) = \frac{4}{7}$

5. Substitute $\lambda$ or $\mu$ back into the vector equation to find the position vector of $P$.

Using the expression with $\lambda$:
$\overrightarrow{OP} = \left(\frac{4}{7}\right)\mathbf{a} + \frac{1}{2}\left(\frac{4}{7}\right)\mathbf{b} = \frac{4}{7}\mathbf{a} + \frac{2}{7}\mathbf{b}$

1. Find the position vector of $Q$.

From $ON : NB = 2 : 1$, we have $\overrightarrow{ON} = \frac{2}{3}\mathbf{b}$.
Since $Q$ lies on $AN$ produced and $AN : NQ = 2 : 1$, the point $N$ divides the line segment $AQ$ internally in the ratio $2 : 1$.
By the ratio theorem:
$\overrightarrow{ON} = \frac{1\overrightarrow{OA} + 2\overrightarrow{OQ}}{2+1}$
$\frac{2}{3}\mathbf{b} = \frac{\mathbf{a} + 2\overrightarrow{OQ}}{3}$
$2\mathbf{b} = \mathbf{a} + 2\overrightarrow{OQ}$
$2\overrightarrow{OQ} = 2\mathbf{b} - \mathbf{a}$
$\overrightarrow{OQ} = \mathbf{b} - \frac{1}{2}\mathbf{a}$

2. Find displacement vectors $\overrightarrow{BQ}$ and $\overrightarrow{BC}$.

$\overrightarrow{BQ} = \overrightarrow{OQ} - \overrightarrow{OB} = \left(\mathbf{b} - \frac{1}{2}\mathbf{a}\right) - \mathbf{b} = -\frac{1}{2}\mathbf{a}$
From the parallelogram $OACB$, $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB} = \mathbf{a} + \mathbf{b}$.
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{a} + \mathbf{b}) - \mathbf{b} = \mathbf{a}$

3. Determine if $Q$, $B$ and $C$ are collinear by finding a scalar relationship.

$\overrightarrow{BC} = \mathbf{a} = -2\left(-\frac{1}{2}\mathbf{a}\right) = -2\overrightarrow{BQ}$
Since $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{BQ}$, the line segments $BC$ and $BQ$ are parallel.
Because point $B$ is common to both segments, $Q$, $B$ and $C$ are collinear.