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

\(\overrightarrow{AB} = (-4 - 4)\mathbf{i} + (3 + 4)\mathbf{j} + (-4 - 1)\mathbf{k} = -8\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (4 - 4)\mathbf{i} + (-1 + 4)\mathbf{j} + (-2 - 1)\mathbf{k} = 0\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\)

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

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 7 & -5 \\ 0 & 3 & -3 \end{vmatrix} = \mathbf{i}(7(-3) - (-5)(3)) - \mathbf{j}(-8(-3) - (-5)(0)) + \mathbf{k}(-8(3) - 7(0))\)

\(= \mathbf{i}(-21 + 15) - \mathbf{j}(24) + \mathbf{k}(-24) = -6\mathbf{i} - 24\mathbf{j} - 24\mathbf{k}\)

Normalizing gives \(\mathbf{n} = \mathbf{i} + 4\mathbf{j} + 4\mathbf{k}\).

Using point \(A(4, -4, 1)\), the equation is:

\(x + 4y + 4z = -8\).

(b) The perpendicular distance from the origin \(O(0, 0, 0)\) to the plane is:

\(\frac{|0 + 4(0) + 4(0) + 8|}{\sqrt{1^2 + 4^2 + 4^2}} = \frac{8}{\sqrt{33}} \approx 1.39\).

(c) The line \(OD\) has the equation \(\mathbf{r} = t(2\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\).

Substitute into the plane equation:

\(2t + 12t - 12t = -8\)

\(2t = -8\)

\(t = -4\)

Coordinates of intersection: \((-8, -12, 12)\).