1. **State the problem:** Given vectors \(\overrightarrow{PA} = \begin{pmatrix} -6 \\ -8 \\ -6 \end{pmatrix}\) and \(\overrightarrow{PN} = \begin{pmatrix} 6 \\ 2 \\ -6 \end{pmatrix}\), find any relevant information such as the vector \(\overrightarrow{AN}\), distance between points, or verify the positions on the 3D prism. 2. **Find vector \(\overrightarrow{AN}\):** Since \(\overrightarrow{PA} = \overrightarrow{P} - \overrightarrow{A}\) and \(\overrightarrow{PN} = \overrightarrow{P} - \overrightarrow{N}\), then $$\overrightarrow{PN} - \overrightarrow{PA} = (\overrightarrow{P} - \overrightarrow{N}) - (\overrightarrow{P} - \overrightarrow{A}) = \overrightarrow{A} - \overrightarrow{N} = - \overrightarrow{AN}.$$ Therefore, $$-\overrightarrow{AN} = \overrightarrow{PN} - \overrightarrow{PA} = \begin{pmatrix}6 \\ 2 \\ -6\end{pmatrix} - \begin{pmatrix}-6 \\ -8 \\ -6\end{pmatrix} = \begin{pmatrix}6+6 \\ 2+8 \\ -6+6\end{pmatrix} = \begin{pmatrix}12 \\ 10 \\ 0\end{pmatrix}.$$ Thus, $$\overrightarrow{AN} = - \begin{pmatrix} 12 \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} -12 \\ -10 \\ 0 \end{pmatrix}.$$ 3. **Interpretation:** This means the vector from point \(A\) to point \(N\) lies in the plane spanned by vectors \(i\) and \(j\) (since the \(k\) component is zero), consistent with \(N\) being a midpoint on base \(DC\) which is horizontal. 4. **Length of \(\overrightarrow{AN}\):** To find the distance between \(A\) and \(N\), compute the magnitude: $$|\overrightarrow{AN}| = \sqrt{(-12)^2 + (-10)^2 + 0^2} = \sqrt{144 + 100} = \sqrt{244} = 2 \sqrt{61} \approx 15.62.$$ 5. **Summary:** The vector \(\overrightarrow{AN}\) is \(\begin{pmatrix} -12 \\ -10 \\ 0 \end{pmatrix}\) with length approximately 15.62 cm, which matches the expected geometry of the base of the prism.