(a) To find the equation of the plane OAB, we need a normal vector. The vectors \(\overrightarrow{OA} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\) are in the plane. The cross product \(\overrightarrow{OA} \times \overrightarrow{OB}\) gives the normal vector \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\). The equation of the plane is \(\mathbf{r} \cdot \mathbf{n} = 0\).

(b) The perpendicular distance from the origin to the plane \(\Pi\) is given by \(\frac{|1|}{\sqrt{1^2 + (-3)^2 + (-2)^2}} = \frac{1}{\sqrt{14}}\).

(c) The angle between the planes is found using the dot product of their normals. The normal to OAB is \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) and for \(\Pi\) is \(\begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}\). The cosine of the angle \(\alpha\) is \(\cos \alpha = \frac{\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}}{\sqrt{3} \sqrt{14}} = \frac{6}{\sqrt{42}}\). The angle is \(22.2^\circ\).

(d) To find the common perpendicular, we first find the direction vector by solving the system of equations derived from the dot product conditions. The direction vector is \(\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\). The position vector is \(\begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix}\). Thus, the equation is \(\mathbf{r} = \begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\).