The parallelepiped satisfies

$$\begin{aligned}
& \mathrm { AB } = 2 , \quad \mathrm { AD } = 3 , \quad \mathrm { AE } = 1 \\
& \angle \mathrm { BAD } = 60 ^ { \circ } , \quad \angle \mathrm { BAE } = 90 ^ { \circ } , \quad \angle \mathrm { DAE } = 120 ^ { \circ }
\end{aligned}$$

and M is the midpoint of edge GH. Let us take points P and Q on edges BF and DH, respectively, such that the four points A, P, M, and Q are on the same plane. We are to find the points P and Q which maximize the length of the line segment PQ.\\
(1) Setting $\vec { a } = \overrightarrow { \mathrm { AB } } , \vec { b } = \overrightarrow { \mathrm { AD } }$ and $\vec { c } = \overrightarrow { \mathrm { AE } }$, the inner products of these vectors are

$$\vec { a } \cdot \vec { b } = \mathbf { A } , \quad \vec { b } \cdot \vec { c } = - \frac { \mathbf { B } } { \mathbf{C} } , \quad \vec { c } \cdot \vec { a } = \mathbf { D } .$$

(2) Let $s$ and $t$ satisfy $0 \leqq s \leqq 1,0 \leqq t \leqq 1$, and set $\mathrm { BP } : \mathrm { PF } = s : ( 1 - s )$, $\mathrm { DQ } : \mathrm { QH } = t : ( 1 - t )$. Since the four points A, P, M and Q are on the same plane, there exist two real numbers $\alpha$ and $\beta$ such that

$$\overrightarrow { \mathrm { AM } } = \alpha \overrightarrow { \mathrm { AP } } + \beta \overrightarrow { \mathrm { AQ } } .$$

Hence $s$ and $t$ satisfy

$$s = \mathbf { E } ( \mathbf { F } - t ) .$$

Then $| \overrightarrow { \mathrm { PQ } } |$ may be expressed in terms of $t$ as

$$| \overrightarrow { \mathrm { PQ } } | ^ { 2 } = \mathbf { G } t ^ { 2 } - \mathbf { H I } t + \mathbf { J K } .$$

Hence the length of segment PQ is maximized when $\square$ L. Here, for $\square$ L choose the correct answer from among choices (0) $\sim$ (5) below.\\
(0) $s = 0 , \quad t = 1$\\
(1) $s = 0 , \quad t = \frac { 1 } { 2 }$\\
(2) $s = \frac { 1 } { 2 } , t = \frac { 3 } { 4 }$\\
(3) $s = \frac { 2 } { 3 } , t = \frac { 2 } { 3 }$\\
(4) $s = 1 , \quad t = \frac { 1 } { 2 }$\\
(5) $s = 1 , \quad t = \frac { 2 } { 3 }$