A rectangular parallelepiped has edges of length 1, 2, and 3 units extending from vertex $A$. The endpoints of these edges, different from $A$, form a triangle. How far is point $A$ from the plane of the triangle?
Let the endpoints of the edges starting from vertex $A$ be denoted by $B$, $C$, and $D$. Let the distance from point $A$ to the plane $BCD$ be $d$. Using the Pythagorean theorem, we can determine the lengths of segments $BC$, $DC$, and $BD$: $BC=\sqrt{5}$, $DC=\sqrt{13}$, $BD=\sqrt{10}$. Let's write down the volume of the tetrahedron $ABCD$ in two ways. First, let the base be triangle $ACD$, then the height is $AB=1$ and $V=\frac{3 \cdot 1}{3}=1$ cubic unit.

![](https://cdn.mathpix.com/cropped/2024_05_02_6ce9020aa9f5d7a6f46bg-1.jpg?height=341&width=463&top_left_y=366&top_left_x=820)

Second, let the base be triangle $BCD$, and denote its area by $T$. Then the height is the distance we are looking for, and $V=\frac{T \cdot d}{3}$. From the equality of the two volumes:

$$
d=\frac{3}{T} \text {. }
$$

Let's calculate the area of triangle $BCD$ using the known area formula:

$$
T=\frac{d \cdot c \cdot \sin \beta}{2}
$$

For this, we first need to determine the value of $\sin \beta$.

Let's write down the cosine rule in triangle $BCD$:

$$
13=5+10-2 \cdot \sqrt{5} \cdot \sqrt{10} \cdot \cos \beta
$$

From this, $\cos \beta=\frac{1}{\sqrt{50}}$. Thus, $\beta0$ and from the Pythagorean identity,

![](https://cdn.mathpix.com/cropped/2024_05_02_6ce9020aa9f5d7a6f46bg-1.jpg?height=268&width=461&top_left_y=1460&top_left_x=821)

$$
\sin \beta=\sqrt{1-\cos ^{2} \beta}=\sqrt{1-\frac{1}{50}}=\sqrt{\frac{49}{50}}
$$

Substituting this into the area formula,

$$
T=\frac{\sqrt{5} \cdot \sqrt{10} \cdot \sqrt{\frac{49}{50}}}{2}=\frac{7}{2}
$$

Substituting this into (1), we get that the distance we are looking for is $d=\frac{3}{\frac{7}{2}}=\frac{6}{7}$ units.