12.6. Consider the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, where $m\left(\angle A_{1} A D\right)=m\left(\angle A_{1} A B\right)=m(\angle D A B)=60^{\circ}$, and $C_{1} A_{1}=\sqrt{7} \mathrm{~cm}, C_{1} B=\sqrt{13} \mathrm{~cm}, C_{1} D=\sqrt{19} \mathrm{~cm}$. Determine the distance from point $A$ to the plane $A_{1} B D$.
## Solution:

Applying the Law of Cosines, we obtain the system

$$
\left\{\begin{array}{c}
A B^{2}+A D^{2}+A B \cdot A D=7 \\
A D^{2}+A_{1} A^{2}+A D \cdot A_{1} A=13 \\
A B^{2}+A_{1} A^{2}+A B \cdot A_{1} A=19
\end{array}\right.
$$

Solving the system

$$
\left\{\begin{array}{c}
A B^{2}+A D^{2}+A B \cdot A D=7 \\
A D^{2}+A_{1} A^{2}+A D \cdot A_{1} A=13 \\
A B^{2}+A_{1} A^{2}+A B \cdot A_{1} A=19
\end{array}\right.
$$

we get $A D=1 \mathrm{~cm}, A B=2 \mathrm{~cm}, A_{1} A=3 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_30675c06d88794a44711g-2.jpg?height=949&width=1268&top_left_y=398&top_left_x=431)

Let $O$ be the projection of point $A_{1}$ onto the plane $A B D$, and $O P \perp A B, P \in(A B)$.

Since $m\left(\angle A_{1} A D\right)=m\left(\angle A_{1} A B\right)=m(\angle D A B)=60^{\circ}$, we get that $m(\angle O A P)=30^{\circ}$.

Then $A P=\frac{3}{2} \mathrm{~cm}, A O=\sqrt{3} \mathrm{~cm}, A_{1} O=\sqrt{6} \mathrm{~cm}$.

Then $\mathcal{V}_{A_{1} A B D}=\frac{1}{3} A_{\triangle A B D} \cdot A_{1} O=\frac{\sqrt{2}}{2}\left(\mathrm{~cm}^{3}\right)$

Applying the Law of Cosines, we get $A_{1} D=A_{1} B=\sqrt{7}(\mathrm{~cm}), B D=\sqrt{3}(\mathrm{~cm})$.

Then $\mathcal{V}_{A_{1} A B D}=\frac{1}{3} A_{\triangle A_{1} B D} \cdot A K=\frac{5 \sqrt{3}}{12} \cdot A K$, where $K$ is the projection of point $A$ onto the plane $A_{1} B D$.

We get $A K=\frac{2 \sqrt{6}}{5} \mathrm{~cm}$.