Given points P(1, 3), Q(-2, 5), R(1, 7), and S(x, y), calculate vectors PQ, QR, PR and find the coordinates of S if PQRS is a parallelogram.

To find the vector from point A to point B, subtract the coordinates of A from the coordinates of B.

$$\vec{PQ} = \begin{pmatrix} x_Q - x_P \\ y_Q - y_P \end{pmatrix} = \begin{pmatrix} -2 - 1 \\ 5 - 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$$

$$\vec{PR} = \begin{pmatrix} x_R - x_P \\ y_R - y_P \end{pmatrix} = \begin{pmatrix} 1 - 1 \\ 7 - 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}$$

The vectors are:

$$\boxed{\vec{PQ} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}, \vec{QR} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}, \vec{PR} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}}$$

In a parallelogram PQRS, the opposite sides are parallel and equal in length. Therefore, $\vec{PS} = \vec{QR}$.
We have P(1, 3) and S(x, y).

$$\vec{PS} = \begin{pmatrix} x - x_P \\ y - y_P \end{pmatrix} = \begin{pmatrix} x - 1 \\ y - 3 \end{pmatrix}$$

From Step 2, we know $\vec{QR} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$.

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

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 S are $(4, 5)$.

$$\boxed{S = (4, 5)}$$