In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, with an edge length of 6, points $M$ and $N$ are the midpoints of edges $A B$ and $B_{1} C_{1}$, respectively, and point $K$ is located on edge $D C$ such that

$D K=2 K C$. Find

a) the distance from point $N$ to line $A K$;

b) the distance between lines $M N$ and $A K$;

c) the distance from point $A_{1}$ to the plane of triangle $M N K$.
![](https://cdn.mathpix.com/cropped/2024_05_06_11215e94233515718ac4g-71.jpg?height=986&width=1993&top_left_y=1099&top_left_x=41)

a) Let $P$ be the orthogonal projection of point $N$ onto the base plane $A B C D$, and $Q$ be the foot of the perpendicular dropped from point $P$ to line $A K$. By the theorem of three perpendiculars, $N Q \perp A K$, so the distance from point $N$ to line $A K$ is the length of segment $N Q$.

Extend $A K$ and $B C$ until they intersect at point $E$. From the similarity of triangles $K C E$ and $K D A$, we find that $C E = A D \cdot \frac{K C}{D K} = 3$.

Let $\angle A E B = \alpha$. Then $\operatorname{ctg} \alpha = \frac{B E}{A B} = \frac{3}{2}$, $\sin \alpha = \frac{1}{\sqrt{1 + \operatorname{ctg}^2 \alpha}} = \frac{2}{\sqrt{13}}$. From the right triangle $P Q E$, we find that $P Q = P E \sin \alpha = 6 \cdot \frac{2}{\sqrt{13}} = \frac{12}{\sqrt{13}}$. Therefore, $N Q = \sqrt{N P^2 + P Q^2} = \sqrt{6^2 + \left(\frac{12}{\sqrt{13}}\right)^2} = \frac{6 \sqrt{17}}{\sqrt{13}}$.

b) Through point $M$, draw a line parallel to $A K$. Let this line intersect $B C$ at point $F$. Then the angle between lines $M N$ and $A K$ is equal to the angle between lines $M N$ and $M F$. From the right triangle $M B F$, we find that $B F = B M \operatorname{ctg} \alpha = \frac{9}{2}$, $M F = \frac{B M}{\sin \alpha} = \frac{3 \sqrt{13}}{2}$.

Thus, $P F = B F - B P = \frac{9}{2} - 3 = \frac{3}{2}$, $N F = \sqrt{N P^2 + P F^2} = \sqrt{6^2 + \left(\frac{3}{2}\right)^2} = \frac{3 \sqrt{17}}{2}$.

Moreover, $P M = B M \sqrt{2} = 3 \sqrt{2}$, $M N = \sqrt{N P^2 + P M^2} = \sqrt{6^2 + (3 \sqrt{2})^2} = 3 \sqrt{6}$.

Consider triangle $M N F$. Let $\angle N M F = \varphi$. By the cosine rule,

$\cos \varphi = \frac{M N^2 + M F^2 - N F^2}{2 M N \cdot M F} = \frac{(3 \sqrt{6})^2 + \left(\frac{3 \sqrt{13}}{2}\right)^2 - \left(\frac{3 \sqrt{17}}{2}\right)^2}{2 \cdot 3 \sqrt{6} \cdot \frac{3 \sqrt{13}}{2}} = \frac{5}{\sqrt{78}}$,

$\sin \varphi = \sqrt{1 - \frac{25}{78}} = \sqrt{\frac{53}{78}}$.

From the right triangle $A D K$, we find that $A K^2 = A D^2 + D K^2 = 36 + 16 = 52$.

Let $V$ be the volume of tetrahedron $A K M N$. Then, on one hand, $V = \frac{1}{3} S_{A K M} \cdot N P = \frac{1}{3} \cdot \frac{1}{2} A M \cdot A D \cdot N P = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 \cdot 6 \cdot 6 = 18$, and on the other hand, if $d$ is the desired distance between lines $A K$ and $M N$, then

$V = \frac{1}{6} A K \cdot M N \cdot d \sin \varphi = \frac{1}{6} \cdot 2 \sqrt{13} \cdot 3 \sqrt{6} \cdot d \cdot \sqrt{\frac{53}{78}} = d \sqrt{53}$.

From the equation $d \sqrt{53} = 18$, we find $d$.

c) Extend segment $M K$ until it intersects line $B C$ at point $H$. From the similarity of triangles $K C H$ and $M B H$, we find that $\frac{2}{3} = \frac{K C}{M B} = \frac{C H}{B H} = \frac{C H}{C H + B C} = \frac{C H}{C H + 6}$, from which $C H = 12$.

Choose a rectangular coordinate system, taking point $C$ as the origin, directing the $C X$ axis along ray $C D$, the $C Y$ axis along ray $C B$, and the $C Z$ axis along ray $C C_{1}$. Find the coordinates of the points we need: $C(0,0,0)$, $K(2,0,0)$, $A(6,6,6)$, $H(0,-12,0)$, $N(0,3,6)$. The equation of plane $M N K$ is $\frac{x}{2} - \frac{y}{12} + \frac{z}{c} = 1$ (the equation of a plane in intercept form), where $(0,0,c)$ are the coordinates of the point where plane $M N K$ intersects the $C Z$ axis. Substituting the coordinates of point $N$ into this equation, we find that $c = \frac{24}{5}$. Therefore, the equation of plane $A B C$ is $-12 x - 2 y + 5 z - 24 = 0$.

The desired distance from point $A_{1}$ to plane $M N K$ is $\frac{|12 \cdot 6 - 2 \cdot 6 + 5 \cdot 6 - 24|}{\sqrt{12^2 + 2^2 + 5^2}} = \frac{66}{\sqrt{173}}$.

## Answer

$6 \sqrt{\frac{17}{13}}, \frac{18}{\sqrt{53}}, \frac{66}{\sqrt{173}}$