(a) To find the obtuse angle between \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\), calculate the scalar product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3)(1) + (-1)(2) + (2)(-3) = 3 - 2 - 6 = -5\)

Calculate the magnitudes:

\(|\overrightarrow{OA}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14}\)

\(|\overrightarrow{OB}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}\)

Use the cosine formula:

\(\cos \theta = \frac{-5}{\sqrt{14} \times \sqrt{14}} = \frac{-5}{14}\)

\(\theta = \cos^{-1}\left(\frac{-5}{14}\right) \approx 110.9^\circ \text{ or } 1.94 \text{ radians}\)

(b) The vector equation for the line \(l\) through points \(A\) and \(B\) is:

\(\mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \mu_1 \begin{pmatrix} 1 - 3 \\ 2 + 1 \\ -3 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \mu_1 \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}\)

(c) For the line through \(C\) and \(D\):

\(\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 5 - 1 \\ -6 + 2 \\ 11 - 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix}\)

Equate components and solve for \(\lambda\) and \(\mu_1\):

\(3 + 2\mu_1 = 1 + 4\lambda\)

\(-1 - 3\mu_1 = -2 - 4\lambda\)

\(2 + 5\mu_1 = 5 + 6\lambda\)

Solve to find \(\mu_1 = 3\) and \(\lambda = 2\).

Substitute back to find the position vector:

\(\mathbf{r} = \begin{pmatrix} 9 \\ -10 \\ 17 \end{pmatrix}\)