令 $\overset{\large\rightharpoonup}{a}=\overset{\large\rightharpoonup}{AB}$，$\overset{\large\rightharpoonup}{b}=\overset{\large\rightharpoonup}{AD}$，$\overset{\large\rightharpoonup}{p}=\overset{\large\rightharpoonup}{AP}$。平行六面體體積 $T=(\overset{\large\rightharpoonup}{a}\times\overset{\large\rightharpoonup}{b})\cdot\overset{\large\rightharpoonup}{p}$。
由向量三重積恆等式：
$$(\overset{\large\rightharpoonup}{b}\times\overset{\large\rightharpoonup}{p})\times(\overset{\large\rightharpoonup}{p}\times\overset{\large\rightharpoonup}{a})
=\bigl((\overset{\large\rightharpoonup}{b}\times\overset{\large\rightharpoonup}{p})\cdot\overset{\large\rightharpoonup}{a}\bigr)\,\overset{\large\rightharpoonup}{p}
=T\,\overset{\large\rightharpoonup}{p}$$
題目給定 $\overset{\large\rightharpoonup}{b}\times\overset{\large\rightharpoonup}{p}=(-2,0,-4)$，$\overset{\large\rightharpoonup}{p}\times\overset{\large\rightharpoonup}{a}=(6,-10,-8)$。計算兩者外積：
\begin{aligned}
T\overset{\large\rightharpoonup}{p}
&=(-2,0,-4)\times(6,-10,-8)\\
&=(0\cdot(-8)-(-4)\cdot(-10),\;(-4)\cdot6-(-2)\cdot(-8),\;(-2)\cdot(-10)-0\cdot6)\\
&=(-40,-40,20)
\end{aligned}
其長度 $|T|\cdot|\overset{\large\rightharpoonup}{p}|=\sqrt{(-40)^2+(-40)^2+20^2}=60$。因為 $\overline{AP}=|\overset{\large\rightharpoonup}{p}|=6$，所以 $|T|=\frac{60}{6}=10$，平行六面體體積為 $10$。
（以下取 $T=10$ 的右手系方向，求各頂點向量）
令 $\overset{\large\rightharpoonup}{u}=\overset{\large\rightharpoonup}{a}\times\overset{\large\rightharpoonup}{b}$，$\overset{\large\rightharpoonup}{v}=\overset{\large\rightharpoonup}{b}\times\overset{\large\rightharpoonup}{p}=(-2,0,-4)$，$\overset{\large\rightharpoonup}{w}=\overset{\large\rightharpoonup}{p}\times\overset{\large\rightharpoonup}{a}=(6,-10,-8)$。
由三重積外積的逆關係：
\begin{aligned}
\overset{\large\rightharpoonup}{w}\times\overset{\large\rightharpoonup}{u}&=T\overset{\large\rightharpoonup}{a}\\
\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}&=T\overset{\large\rightharpoonup}{b}\\
\overset{\large\rightharpoonup}{v}\times\overset{\large\rightharpoonup}{w}&=T\overset{\large\rightharpoonup}{p}
\end{aligned}
其中 $\overset{\large\rightharpoonup}{p}=(\overset{\large\rightharpoonup}{v}\times\overset{\large\rightharpoonup}{w})/T=(-40,-40,20)/10=(-4,-4,2)$（$|\overset{\large\rightharpoonup}{p}|=6$，驗證無誤）。
取與三外積相容的 $\overset{\large\rightharpoonup}{u}=(-5,5,5)$（可由 $\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{p}=T$ 及其他正交條件定出），則
\begin{aligned}
\overset{\large\rightharpoonup}{a}&=\frac{\overset{\large\rightharpoonup}{w}\times\overset{\large\rightharpoonup}{u}}{T}
=\frac{(6,-10,-8)\times(-5,5,5)}{10}=(-1,1,-2)\\[4pt]
\overset{\large\rightharpoonup}{b}&=\frac{\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}}{T}
=\frac{(-5,5,5)\times(-2,0,-4)}{10}=(-2,-3,1)
\end{aligned}
平行六面體以 $A$ 為原點的八個頂點向量為 $\overset{\large\rightharpoonup}{0},\overset{\large\rightharpoonup}{a},\overset{\large\rightharpoonup}{b},\overset{\large\rightharpoonup}{p},
\overset{\large\rightharpoonup}{a}+\overset{\large\rightharpoonup}{b},\overset{\large\rightharpoonup}{a}+\overset{\large\rightharpoonup}{p},
\overset{\large\rightharpoonup}{b}+\overset{\large\rightharpoonup}{p},\overset{\large\rightharpoonup}{a}+\overset{\large\rightharpoonup}{b}+\overset{\large\rightharpoonup}{p}$。
各頂點到 $A$ 的距離平方為：
$$0,\;6,\;14,\;36,\;14,\;34,\;94,\;86$$
其中最大值為 $94$，故最長距離為 $\sqrt{94}$。
故體積為 $10$，最長距離為 $\sqrt{94}$。
