(a) To find the equation of the plane ABC, we first determine the direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 6 \\ 4 \\ -2 \end{pmatrix}\)

Substituting point \(A(1, 0, -2)\) into the plane equation \(3x + 2y - z = d\), we find \(d = 5\).

(b) The normal to the plane ABD is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = \begin{pmatrix} 2t + 4 \\ 0 \\ 0 \end{pmatrix}\)

The angle \(\theta\) between the planes is given by:

\(\cos \theta = \frac{\left( \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right)}{\sqrt{14} \cdot 1}\)

\(\theta = 36.7^\circ\).

(c) To find \(t\) such that the shortest distance between lines AB and CD is \(\sqrt{2}\), we use the formula for the perpendicular distance between skew lines:

\(\frac{1}{\sqrt{t^2 - 2t + 6}} \left| \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \right| = \frac{t + 2}{\sqrt{t^2 - 2t + 6}}\)

Setting this equal to \(\sqrt{2}\) and solving gives:

\(t^2 - 8t + 8 = 0 \Rightarrow t = 2(2 \pm \sqrt{2})\)

Thus, \(t = 6.83\) and \(t = 1.17\).