## Solution.

1) If $A, B, C$ are the locations of the ships, then triangle $A B C$ is isosceles with angles $B=120^{\circ}, A=C=30^{\circ}$ and $A B=B C=20$ km. Then the desired locus of points in the plane of triangle $A B C$ is the point $O$ - the center of the circumscribed circle

![](https://cdn.mathpix.com/cropped/2024_06_04_c790142104ba3f7af959g-17.jpg?height=474&width=700&top_left_y=603&top_left_x=1249)
as the intersection of the perpendicular bisectors of triangle $A B C$. (1 point)

2) The radii of this circle $A O=B O=C O=R=($ to be proven! $)=A B=B C=20$ km.
3) Then in space $R^{3}$, the desired locus of points is a part of the line passing through point $O$ perpendicular to the plane $A B C$.
4) The final locus of points is determined by two parts of this line:

- the underwater part (where a submarine or underwater buoy can be placed), which is determined by the depth of the sea.

(1 point)(7 points)

- the above-water part up to a height $h=\sqrt{30^{2}-20^{2}}=10 \sqrt{5}$ km, where a drone can be placed.

(+1 point)(7 points)
3. In the sea off the coast of Kamchatka, three suspicious fishing vessels are regularly detected, traditionally located at the vertices of an isosceles triangle with an angle of $120^{\circ}$ and side lengths of 20 km. Find the geometric locus of points (GML) in space where technical control means (with a range of no more than 30 km) for the illegal harvesting of Kamchatka crab can be placed, so that all points of this GML are equidistant from the vessels - potential violators of environmental legislation.