Find the parametric equation of line AB and the Cartesian equation of line d.

The coordinates of points $A$ and $B$ are $A(1,0)$ and $B(-2,3)$.
First, find the direction vector $\vec{AB}$.

$$\vec{AB} = (x_B - x_A, y_B - y_A) = (-2 - 1, 3 - 0) = (-3, 3)$$

The parametric equation of a line passing through a point $(x_0, y_0)$ with a direction vector $(u,v)$ is given by:

$$\begin{cases} x = x_0 + t \cdot u \\ y = y_0 + t \cdot v \end{cases}$$

Using point $A(1,0)$ and the direction vector $\vec{AB}=(-3,3)$, and letting the parameter be $t$:

$$\begin{cases} x = 1 + t(-3) \\ y = 0 + t(3) \end{cases}$$ $$\begin{cases} x = 1 - 3t \\ y = 3t \end{cases}$$

The student used $m$ as the parameter, which is also acceptable.

$$\boxed{(AB): \begin{cases} x = 1 - 3m \\ y = 3m \end{cases}}$$

The line $(d)$ is given by the parametric equations:

$$\begin{cases} x = m+1 \quad (1) \\ y = -2m+3 \quad (2) \end{cases}$$

From equation (1), express $m$ in terms of $x$:

$$m = x-1$$

Substitute this expression for $m$ into equation (2):

$$y = -2(x-1)+3$$ $$y = -2x+2+3$$ $$y = -2x+5$$

To write the equation in the standard Cartesian form $ax+by+c=0$, move all terms to one side:

$$2x+y-5 = 0$$

The student's work for this part is correct.

$$\boxed{(d): 2x+y-5=0}$$