12.287. The edges of a rectangular parallelepiped are in the ratio $3: 4: 12$. A diagonal section is made through the larger edge. Find the sine of the angle between the plane of this section and a diagonal of the parallelepiped not lying in it.
## Solution.

Let $O$ and $O_{1}$ be the points of intersection of the diagonals of the bases of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.150), $A_{1} B_{1}=3 ; B_{1} C_{1}=$ $=4 ; B B_{1}=12$.

Then $A_{1} C_{1}=\sqrt{A_{1} B_{1}^{2}+B_{1} C_{1}^{2}}=5$;

$$
B_{1} D=\sqrt{A_{1} B_{1}^{2}+B_{1} C_{1}^{2}+B B_{1}^{2}}=13
$$

Draw a perpendicular $B_{1} F$ to $A_{1} C_{1}$ in the plane of the rectangle $A_{1} B_{1} C_{1} D_{1}$. Since the plane of the diagonal section $A A_{1} C_{1} C$ is perpendicular to the planes of the bases, $B_{1} F$ is perpendicular to the plane of the section, and the line $F E$ ( $F$ is the intersection point of $B_{1} D$ and the section - the midpoint of $O O_{1}$ ) is the projection of $B_{1} D$ onto the plane of the section, and $\angle B_{1} E F$ is the required angle between the plane of the section $A A_{1} C_{1} C$ and $B_{1} D$.

$$
S_{\triangle A_{1} B_{1} C_{1}}=\frac{1}{2} A_{1} B_{1} \cdot B_{1} C_{1}=\frac{1}{2} A_{1} C_{1} \cdot B_{1} F \Rightarrow B_{1} F=\frac{A_{1} B_{1} \cdot B_{1} C_{1}}{A_{1} C_{1}}=\frac{12}{5}
$$

In $\triangle B_{1} F E\left(\angle B_{1} F E=90^{\circ}\right)$ :

$\sin \angle B_{1} E F=\frac{B_{1} F}{B_{1} E}=\frac{12}{5}: \frac{13}{2}=\frac{24}{65}$.

Answer: $\frac{24}{65}$.