Question 1209329
To solve this problem, we need to consider the transformations that can move point P to each vertex exactly once. 

The value of k in the probability $\frac{k}{11^6}$ can be calculated as follows:

1. There are 6! = 720 ways to arrange the six vertices in a sequence.

2. For each sequence, we need to count the number of ways to choose transformations that achieve that sequence:

   - The first transformation can be any of the 11 that moves P to the first vertex in the sequence.
   - The second transformation must be one of the 2 that moves the current position to the second vertex.
   - Each subsequent transformation also has 2 choices to move to the next vertex.

3. Therefore, for each sequence, there are 11 * 2^5 = 352 ways to choose the transformations.

4. The total number of favorable outcomes is thus 720 * 352 = 253,440.

Therefore, k = 253,440.