The points A(1, 3), B(-2, 5), C(1, 7) and D(x, y) are the vertices of the parallelogram ABCD. Find vector AB, vector BC, vector AC, and the coordinates of D.

To find the vector from point $P(x_1, y_1)$ to $Q(x_2, y_2)$, we use $\vec{PQ} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$.
For $\vec{AB}$, $A(1, 3)$ and $B(-2, 5)$:

$$\vec{AB} = \begin{pmatrix} -2 - 1 \\ 5 - 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$$ $$\boxed{\vec{AB} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}}$$

For $\vec{BC}$, $B(-2, 5)$ and $C(1, 7)$:

$$\vec{BC} = \begin{pmatrix} 1 - (-2) \\ 7 - 5 \end{pmatrix} = \begin{pmatrix} 1 + 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$$ $$\boxed{\vec{BC} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}}$$

For $\vec{AC}$, $A(1, 3)$ and $C(1, 7)$:

$$\vec{AC} = \begin{pmatrix} 1 - 1 \\ 7 - 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}$$ $$\boxed{\vec{AC} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}}$$

Therefore, their vectors are equal. For parallelogram ABCD, $\vec{AD} = \vec{BC}$.
Let $D = (x, y)$.

$$\vec{AD} = \begin{pmatrix} x - 1 \\ y - 3 \end{pmatrix}$$

From part (i), we know $\vec{BC} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$.

$$\begin{pmatrix} x - 1 \\ y - 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$$

Equating the x-components:

$$x - 1 = 3$$ $$x = 3 + 1$$ $$x = 4$$

Equating the y-components:

$$y - 3 = 2$$ $$y = 2 + 3$$ $$y = 5$$

The coordinates of D are $(4, 5)$.

$$\boxed{D(4, 5)}$$