(a) The equation of the plane \(\Pi_1\) can be found by setting up the system of equations from the given vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\)

Solving gives \(y + z = -7\).

(b) To find the line of intersection, find a point common to both planes, e.g., \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\), and the direction vector by solving:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -5 & 3 & 5 \end{vmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\)

The vector equation is \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\).

(c) The perpendicular distance from \(A\) to \(\Pi_1\) is given by:

\(\frac{|a + a - 7 + 7|}{\sqrt{2}} = \sqrt{2}\)

Solving gives \(a = 1\).

(d) The obtuse angle condition gives:

\(\frac{|b + b|}{\sqrt{2((1-b)^2 + 2b^2)}} = \frac{1}{2}\sqrt{2}\)

Solving \(2b = \sqrt{(1-b)^2 + 2b^2}\) gives \(b = -1 + \sqrt{2}\).