(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 \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):
\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = (-5(2) - 5(-5))\mathbf{i} + (0(-5) - (-5)(-5))\mathbf{j} + (0(-5) - (-5)(2))\mathbf{k}\)
\(= -10\mathbf{i} + 25\mathbf{j} + 10\mathbf{k}\)
\(= -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)
Using point \(A(2, 2, 4)\), the equation of the plane is:
\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)
\(-5x + 5y + 2z = 8\)
(b) The perpendicular distance from point \(D(2, 1, 3)\) to the plane is given by:
\(\frac{|(-5)(2) + 5(1) + 2(3) - 8|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} \approx 0.953\)
(c) The shortest distance between the lines AB and CD is found using the vector \(\overrightarrow{CD}\) and the cross product:
\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = -3\mathbf{j} + 4\mathbf{k}\)
\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ 5 & 4 & -1 \end{vmatrix} = (18)\mathbf{i} + (-25)\mathbf{j} + (-10)\mathbf{k}\)
The shortest distance is:
\(\frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{35}{\sqrt{1049}} \approx 1.08\)
