The distances from the vertices of a triangle to a certain plane are 5, 6, and 7. Find the distance from the point of intersection of the medians of this triangle to the same plane. List all possibilities.

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Let $A 1, B 1, C 1, K 1$ and $M 1$ be the orthogonal projections of the vertices $A, B$ and $C$ of triangle $A B C$, the midpoint $K$ of side $A C$, and the point $M$ of intersection of the medians of this triangle onto the plane $\alpha$, respectively, and $A A 1=5$, $B B 1=6$ and $C C 1=7$. If points $A, B$ and $C$ are located on the same side of the plane $\alpha$ (Fig.1), then $K K 1$ is the midline of the rectangular trapezoid $A A 1 C 1 C$ with bases $A A 1$ and $C C 1$. Therefore,

$$
K K 1=\frac{1}{2}(A A 1+C C 1)=\frac{1}{2}(5+7)=6 .
$$

The quadrilateral $B B 1 K 1 K$ is a rectangle, in which $B B 1=K K 1=6$. Points $M$ and $M 1$ lie on its opposite sides $B K$ and $B 1 K 1$, respectively, and $M M 1 \| B B 1$. Therefore, $M M 1=B B 1=6$. If points $A, C$ are located on the same side of the plane $\alpha$, and point $B$ is on the other side (Fig.2), similarly to the previous case, we find that $K K 1=6$. The quadrilateral $B B 1 K K 1$ is a parallelogram, in which $B B 1=K K 1=6$. Points $M$ and $M 1$ lie on its diagonals $B K$ and $B 1 K 1$, respectively, and $M M 1 \| B B 1$ and $\frac{K M}{M B}=\frac{1}{2}$. Let the line $M M 1$ intersect side $B K 1$ at point $D$. Triangle $B M D$ is similar to triangle $B K K 1$ with a coefficient of $\frac{2}{3}$, and triangle $K 1 D M 1$ is similar to triangle $K 1 B B 1$ with a coefficient of $\frac{1}{3}$. Therefore,

$$
M M 1=M D-M 1 D=\frac{2}{3} K K 1-\frac{1}{3} B B 1=4-2=2
$$

If points $A$ and $B$ are located on the same side of the plane $\alpha$, and point $C$ is on the other side (Fig.3), then $K K 1$ is the segment connecting the midpoints of the diagonals $A C$ and $A 1 C 1$ of the trapezoid $A A 1 C C 1$ with bases $A A 1=5$ and $C C 1=7$. Therefore,

$$
K K 1=\frac{1}{2}(C C 1-A A 1)=\frac{1}{2}(7-5)=1
$$

and points $C$ and $K$ lie on the same side of the plane $\alpha$, which means points $B$ and $K$ are on opposite sides of this plane. Then the segment $M M 1$ connects points $M$ and $M 1$, lying on the diagonals $B K$ and $B 1 K 1$ of the trapezoid $B B 1 K K 1$ with bases $B B 1=6$ and $K K 1=1$, respectively, and $\frac{K M}{M B}=\frac{1}{2}$ and $M M 1 \| B B 1$. Let the line $M M 1$ intersect side $B 1 K$ at point $E$. Triangle $K M E$ is similar to triangle $K B B 1$ with a coefficient of $\frac{1}{3}$, and triangle $B 1 M 1 E$ is similar to triangle $B 1 K 1 K$ with a coefficient of $\frac{2}{3}$. Therefore,

$$
M M 1=M E-M 1 E=\frac{1}{3} B B 1-\frac{2}{3} K K 1=2-\frac{2}{3}=\frac{4}{3} .
$$

Finally, if points $B$ and $C$ are located on the same side of the plane $\alpha$, and point $A$ is on the other side (Fig.4), then similarly to the previous case, we find that $K K 1=1$ and points $K$ and $B$ are located on the same side of the plane $\alpha$. Then the endpoints of the segment $M M 1$ lie on the lateral sides of the rectangular trapezoid $B B 1 K 1 K$ with bases $B B 1$ and $K K 1$, respectively, and $\frac{K M}{M B}=\frac{1}{2}$ and $M M 1 \| B B 1$. Let the diagonal $B K 1$ intersect the segment $M M 1$ at point $F$. Triangle $B M F$ is similar to triangle $B K K 1$ with a coefficient of $\frac{2}{3}$, and triangle $K 1 M 1 F$ is similar to triangle $K 1 B 1 B$ with a coefficient of $\frac{1}{3}$. Therefore,

$$
M M 1=M F+M 1 F=\frac{2}{3} K K 1+\frac{1}{3} B B 1=\frac{2}{3}+2=\frac{8}{3} .
$$

## Answer

$6 ; 2 ; \frac{4}{3} ; \frac{8}{3}$