(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

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

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

Simplifying gives \(-5x + 5y + 2z = 8\).

(b) The perpendicular distance from point D to the plane is given by:

\(\frac{|8 - (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \cdot (-5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} = 0.953\)

(c) The direction vector \(\overrightarrow{CD}\) is:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 4\mathbf{j} - \mathbf{k}\)

Using the formula for the shortest distance between two skew lines:

\(\frac{|\overrightarrow{AB} \cdot (\overrightarrow{CD} \times \overrightarrow{AC})|}{|\overrightarrow{CD} \times \overrightarrow{AC}|} = \frac{35}{\sqrt{1049}} = 1.08\)