Relative to an origin $O$, the points $C$ and $D$ have position vectors $c=\begin{pmatrix}2\\3\\2\end{pmatrix}\quad\text{and}\quad d=\begin{pmatrix}3\\\alpha\\\beta\end{pmatrix}$ respectively, where $\alpha$ and $\beta$ are real.

1. Identify that the coordinates of $D$ must satisfy the line's Cartesian equations, and solve for $\beta$.

Since $D(3, \alpha, \beta)$ lies on line $l$, its coordinates satisfy the equation $z = 2$.
$\beta = 2$

2. Substitute the $x$ and $y$ coordinates of $D$ into the other part of the line's equation.

Substitute $x = 3$ and $y = \alpha$ into $2 - x = \dfrac{y - 3}{3}$:
$2 - 3 = \frac{\alpha - 3}{3}$

3. Solve the linear equation for $\alpha$.

$-1 = \frac{\alpha - 3}{3}$
$-3 = \alpha - 3$
$\alpha = 0$

1. State the condition for the y-z plane and substitute it into the Cartesian equation of the line.

The equation of the y-z plane is $x = 0$.
Substitute $x = 0$ into the equation of $l$:
$2 - 0 = \frac{y-3}{3}$

2. Solve for the remaining coordinates to find the point of intersection.

$2 = \frac{y-3}{3}$
$y - 3 = 6 \implies y = 9$
From the equation of $l$, we are also given $z = 2$.
Therefore, the point of intersection is $(0, 9, 2)$.

1. Determine the coordinates of $C'$ by applying the reflection property in the y-z plane.

When a point is reflected in the y-z plane (where $x=0$), its x-coordinate is negated while its y and z coordinates remain unchanged.

Given the position vector of $C$ is $\begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}$ (which corresponds to the point $(2, 3, 2)$).

The coordinates of the reflected point $C'$ will be $(-2, 3, 2)$.

Therefore, the position vector of $C'$ is:
$c' = \begin{pmatrix} -2 \\ 3 \\ 2 \end{pmatrix}$