Given p = OA + (1)/(2) OB + 3OC.

Given $p = OA + \frac{1}{2} OB + 3OC$.
Substitute the given vectors:

$$p = \begin{pmatrix} 5 \\ -2 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 2 \\ 4 \end{pmatrix} + 3 \begin{pmatrix} -4 \\ 2 \end{pmatrix}$$

$$p = \begin{pmatrix} 5 \\ -2 \end{pmatrix} + \begin{pmatrix} \frac{1}{2} \times 2 \\ \frac{1}{2} \times 4 \end{pmatrix} + \begin{pmatrix} 3 \times -4 \\ 3 \times 2 \end{pmatrix}$$ $$p = \begin{pmatrix} 5 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} -12 \\ 6 \end{pmatrix}$$

$$p = \begin{pmatrix} 5 + 1 - 12 \\ -2 + 2 + 6 \end{pmatrix}$$ $$p = \begin{pmatrix} 6 - 12 \\ 0 + 6 \end{pmatrix}$$ $$p = \begin{pmatrix} -6 \\ 6 \end{pmatrix}$$

The magnitude of a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2 + y^2}$.

$$|p| = \sqrt{(-6)^2 + 6^2}$$ $$|p| = \sqrt{36 + 36}$$ $$|p| = \sqrt{72}$$

Simplify the square root:

$$|p| = \sqrt{36 \times 2} = 6\sqrt{2}$$

A unit vector is given by $\frac{p}{|p|}$.

$$\text{Unit vector} = \frac{1}{6\sqrt{2}} \begin{pmatrix} -6 \\ 6 \end{pmatrix}$$ $$\text{Unit vector} = \begin{pmatrix} \frac{-6}{6\sqrt{2}} \\ \frac{6}{6\sqrt{2}} \end{pmatrix}$$ $$\text{Unit vector} = \begin{pmatrix} \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$$

Rationalize the denominators:

$$\text{Unit vector} = \begin{pmatrix} \frac{-1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \\ \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \end{pmatrix}$$ $$\text{Unit vector} = \begin{pmatrix} -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \end{pmatrix}$$

The unit vector is $\boxed{\begin{pmatrix} -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \end{pmatrix}}$.