以原点 $O$ 为基准，三个点 $A$、$B$ 和 $C$ 的位置向量分别为 $3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}$、$\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ 和 $\alpha\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}$。

1. Find the position vector $\overrightarrow{OD}$ in terms of $\lambda$.

$\overrightarrow{OD} = \overrightarrow{OA} + \lambda \overrightarrow{OB} = \begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 + \lambda \\ -2 + 4\lambda \\ 5 - 2\lambda \end{pmatrix}$

2. Use the condition for perpendicular vectors to solve for $\lambda$.

For $\overrightarrow{OD}$ to be perpendicular to the $y$-axis, its dot product with the $y$-axis direction vector $\mathbf{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ must be zero (which means its $y$-component must be $0$).

$\begin{pmatrix} 3 + \lambda \\ -2 + 4\lambda \\ 5 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 0$
$-2 + 4\lambda = 0$
$4\lambda = 2 \implies \lambda = \frac{1}{2}$

1. Calculate the value of $\lambda$ and the vector $\overrightarrow{OD}$ from part (b).

From part (b), $\overrightarrow{OD} = \overrightarrow{OA} + \lambda \overrightarrow{OB}$:
$\overrightarrow{OD} = \begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 + \lambda \\ -2 + 4\lambda \\ 5 - 2\lambda \end{pmatrix}$
Since $\overrightarrow{OD}$ is perpendicular to the $y$-axis, its $y$-component must be $0$:
$-2 + 4\lambda = 0 \implies \lambda = \frac{1}{2}$
Substituting $\lambda = \frac{1}{2}$ gives:
$\overrightarrow{OD} = \begin{pmatrix} 7/2 \\ 0 \\ 4 \end{pmatrix}$

2. Evaluate the cross product $\overrightarrow{OA} \times \overrightarrow{OD}$.

$\overrightarrow{OA} \times \overrightarrow{OD} = \begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} \times \begin{pmatrix} 7/2 \\ 0 \\ 4 \end{pmatrix}$
$= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 5 \\ 7/2 & 0 & 4 \end{vmatrix}$
$= \mathbf{i}((-2)(4) - 0) - \mathbf{j}((3)(4) - (5)(\frac{7}{2})) + \mathbf{k}(0 - (-2)(\frac{7}{2}))$
$= \mathbf{i}(-8) - \mathbf{j}(12 - \frac{35}{2}) + \mathbf{k}(7)$
$= \begin{pmatrix} -8 \\ 11/2 \\ 7 \end{pmatrix}$

3. Calculate the exact area of triangle $OAD$.

$\text{Area} = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OD}|$
$= \frac{1}{2} \sqrt{(-8)^2 + \left(\frac{11}{2}\right)^2 + 7^2}$
$= \frac{1}{2} \sqrt{64 + \frac{121}{4} + 49}$
$= \frac{1}{2} \sqrt{113 + \frac{121}{4}}$
$= \frac{1}{2} \sqrt{\frac{573}{4}}$
$= \frac{\sqrt{573}}{4}$