(a) To find the direction vector of \(AB\), calculate \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix}\).

The direction vector of line \(l\) is \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).

Calculate the dot product: \(\begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 2 \times 1 + (-1) \times (-2) + (-3) \times 1 = 2 + 2 - 3 = 1\).

Calculate the magnitudes: \(\| \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix} \| = \sqrt{2^2 + (-1)^2 + (-3)^2} = \sqrt{14}\) and \(\| \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}\).

Use the cosine formula: \(\cos \theta = \frac{1}{\sqrt{6} \sqrt{14}}\).

Find \(\theta\) using inverse cosine: \(\theta = \cos^{-1} \left( \frac{1}{\sqrt{6} \sqrt{14}} \right) \approx 83.7^\circ\) or \(1.46\) radians.

(b) Express \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) in terms of \(\lambda\): \(\overrightarrow{AP} = \begin{pmatrix} 1 + \lambda \\ 1 - 2\lambda \\ \lambda \end{pmatrix}\) and \(\overrightarrow{BP} = \begin{pmatrix} -1 + \lambda \\ 2 - 2\lambda \\ 3 + \lambda \end{pmatrix}\).

Equate the magnitudes: \((1 + \lambda)^2 + (1 - 2\lambda)^2 + \lambda^2 = (-1 + \lambda)^2 + (2 - 2\lambda)^2 + (3 + \lambda)^2\).

Simplify and solve for \(\lambda\): \(83\lambda^2 - 528\lambda + 207 = 0\).

Solving gives \(\lambda = 6\).

Substitute \(\lambda = 6\) into the line equation to find \(P\): \(\mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + 6 \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ -9 \\ 7 \end{pmatrix}\).