Problem: A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: We introduce a Cartesian coordinate system to the diagram.
We put the origin at $A$ . We let the $z$ -components of $B$ $C$ $D$ be positive.
We set the $x$ -axis in a direction such that $B$ is on the $x-O-z$ plane.
The coordinates of $A$ $B$ $C$ are $A = \left( 0, 0, 0 \right)$ $B = \left( x_B, 0 , 2 \right)$ $C = \left( x_C, y_C, 8 \right)$
Because $AB \perp AC$ $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$ .
Thus, \[ x_B x_C + 16 = 0 . \hspace{1cm} (1) \]
Because $AC$ is a diagonal of a face, $AC^2 = 2 AB^2$ .
Thus, \[ x_C^2 + y_C^2 + 8^2 = 2 \left( x_B^2 + 2^2 \right) . \hspace{1cm} (2) \]
Because plane $ABCD$ is perpendicular to plan $P$ $\hat z \cdot \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) = 0$ .
Thus, \[ \begin{vmatrix} 0 & 0 & 1 \\ x_B & 0 & 2 \\ x_C & y_C & 8 \end{vmatrix} = 0 . \hspace{1cm} (3) \]
Jointly solving (1), (2), (3), we get one solution $x_B = 4 \sqrt{2}$ $x_C = - 2 \sqrt{2}$ $y_C = 0$ .
Thus, the side length of the cube is $|AB| = \sqrt{x_B^2 + 2^2} = 6$
Denote by $P$ and $Q$ two vertices such that $AP$ and $AQ$ are two edges, and satisfy the right-hand rule that $\widehat{AB} \times \widehat{AP} = \widehat{AQ}$ .
Now, we compute the coordinates of $P$ and $Q$
Because $|AB| = 6$ , we have $\overrightarrow{AP} \times \overrightarrow{AQ} = 6 \overrightarrow{AB}$ $\overrightarrow{AQ} \times \overrightarrow{AB} = 6 \overrightarrow{AP}$ $\overrightarrow{AB} \times \overrightarrow{AP} = 6 \overrightarrow{AQ}$
Hence, \begin{align*} \begin{bmatrix} \hat i & \hat j & \hat k \\ x_P & y_P & z_P \\ x_Q & y_Q & z_Q \end{bmatrix} & = 6 \left( 4 \sqrt{2} \hat i + 2 \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ x_Q & y_Q & z_Q \\ 4 \sqrt{2} & 0 & 2 \end{vmatrix} & = 6 \left( x_P \hat i + y_P \hat j + z_P \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 \sqrt{2} & 0 & 2 \\ x_P & y_P & z_P \end{vmatrix} & = 6 \left( x_Q \hat i + y_Q \hat j + z_Q \hat k \right) . \end{align*}
By solving these equations, we get $y_P^2 + y_Q^2 = 36 .$
In addition, we have $\overrightarrow{AC} = \overrightarrow{AP} + \overrightarrow{AQ}$ .
Thus, $P = \left( - \sqrt{2} , 3 \sqrt{2} , 4 \right)$ $Q = \left( - \sqrt{2} , - 3 \sqrt{2} , 4 \right)$
Therefore, the volume of the water is \begin{align*} V = & 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ z_B u + z_P v + z_Q w \leq 7 \right\} dw dv du \\ & = 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w \leq 7 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w > 7 \right\} dw dv du . \end{align*}
Define $u' = 1 - u$ $v' = 1 - v$ $w' = 1 - w$ .
Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u' + 4 v' + 4 w' < 3 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u'=0}^1 \left( \int_{v'=0}^1 \int_{w'=0}^1  \mathbf 1 \left\{ v' + w' < \frac{3}{4} - \frac{u'}{2} \right\} dw' dv' \right)  du' \\ & = 6^3 - 6^3 \int_{u'=0}^1  \frac{1}{2} \left( \frac{3}{4} - \frac{u'}{2} \right)^2 du' . \end{align*}
Define $u'' = \frac{3}{4} - \frac{u'}{2}$ .
Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u'' = 1/4}^{3/4} \left( u'' \right)^2 du'' \\ & = 6^3 - 6^3 \frac{1}{3} \left( \left(\frac{3}{4}\right)^3 - \left(\frac{1}{4}\right)^3 \right) \\ & = 216 - \frac{117}{4} \\ & = \frac{747}{4} . \end{align*}
Therefore, the answer is $747 + 4 = \boxed{751}$

Answer:  751