Three distinct vertices are randomly chosen among the vertices of a cube. The probability that they from equilateral triangle is

Total number of ways in which three vertices can be selected from the vertices of a cube ${=}^8{C}_3$.

Here, for the triangle to be equilateral the sides of the triangle can only be along face diagonals as no three edges of the cube can form an equilateral triangle.

Also, for body diagonals, only one body diagonal can be drawn from a vertex which again eliminates the possibility of a triangle.

Now, the three face diagonals chosen must be on any three faces with co-terminus edges as shown in the figure,

Here, the faces $A, B, C$ are co-terminus as they have edges originating from a common point.

For one set of such faces, only one equilateral triangle is possible where the vertices are the end-points of their co-terminus edges other than the origin point.

It can also be observed that from one vertex of the cube, only one set of co-terminus edges can be formed.

As there are  $8$ vertices, total such sets of edges $= 8$ and hence total possible equilateral triangles $=8$.

Now, we know $P\left(Event\right) = \frac{Total favourable cases}{Total possible cases}$.

Hence, $P(Equilateral triangle is formed) = \frac{8}{{ }^8{C}_3} = \frac{1}{7}$

Hence, option $3$ is correct.