Classify acceleration structure using surface area heuristic
You gave the formula as:

$$Cost(cell) = C_t + \frac{S_L}{S_{Cell}} \times N_L \times t_i + \frac{S_R}{S_{Cell}} \times N_R \times t_i$$

(You put $r_i$ at the right-hand end, but I assume that was a typo.) 

There's a more general form of this formula

$$Cost(cell) = C_t + \sum_{k\ \in\ K} \frac{S_k}{S_{Cell}} \times N_k \times t_i$$

where $K$ is the set of all child nodes; that is, to get back to the two-children form you started with, $K = \{L, R\}$.

By generalising the formula like this, you can now calculate the cost for the first possibility the same way as the second two.

You're right that this is a slightly odd question. In practice, your traversal cost is going to be less for a binary tree than a $(&gt; 2)$-ary tree.


  
  Does this ranking change if we change the spheres with a (scaled) tea-pot that still fits within the bounding volume?
  


To answer this, you'll need to consider how this affects the surface area, and how this changes each score in turn.


  
  Can this ranking depend on the position of the camera, when we only    consider visibility rays (so no shadow rays, for example)? You can assume you're working with an orthographic camera.
  


There's two things to consider here.

1) Does any part of the heuristic depend on the position of the camera? Can moving the camera change any of the numbers that feed into the score?

2) Moving the camera can change how the boxes overlap. Remember that rays that pass through the box need to be tested against the box behind, so needless overlaps make the structure worse.

I think the point is that the heuristic is only a general rule. It's not hard to construct an example with two proposals a and b, where a scores much better but b is actually a better split for a particular camera, and this last part of the question is trying to get you to think about this.