Calculate the area of a parallelogram with adjacent sides represented by vectors (2i - 3k) and (4j - 2k).

Let $\vec{a}$ and $\vec{b}$ be the adjacent sides.

$$\vec{a} = 2\hat{i} - 3\hat{k} = 2\hat{i} + 0\hat{j} - 3\hat{k}$$ $$\vec{b} = 4\hat{j} - 2\hat{k} = 0\hat{i} + 4\hat{j} - 2\hat{k}$$

The area of the parallelogram is given by the magnitude of the cross product of its adjacent sides.

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -3 \\ 0 & 4 & -2 \end{vmatrix}$$ $$\vec{a} \times \vec{b} = \hat{i}((0)(-2) - (-3)(4)) - \hat{j}((2)(-2) - (-3)(0)) + \hat{k}((2)(4) - (0)(0))$$ $$\vec{a} \times \vec{b} = \hat{i}(0 - (-12)) - \hat{j}(-4 - 0) + \hat{k}(8 - 0)$$ $$\vec{a} \times \vec{b} = 12\hat{i} + 4\hat{j} + 8\hat{k}$$

The area of the parallelogram is $||\vec{a} \times \vec{b}||$.

$$||\vec{a} \times \vec{b}|| = \sqrt{(12)^2 + (4)^2 + (8)^2}$$ $$||\vec{a} \times \vec{b}|| = \sqrt{144 + 16 + 64}$$ $$||\vec{a} \times \vec{b}|| = \sqrt{224}$$

To simplify $\sqrt{224}$, find the largest perfect square factor of 224.
$224 = 16 \times 14$

$$||\vec{a} \times \vec{b}|| = \sqrt{16 \times 14}$$ $$||\vec{a} \times \vec{b}|| = \sqrt{16} \times \sqrt{14}$$ $$||\vec{a} \times \vec{b}|| = 4\sqrt{14}$$

The calculated area is $4\sqrt{14}$. This matches option (b).

The final answer is $\text{(b)} 4\sqrt{14}$.