'Twas the night before Christmas and throughout the net,
Not a question was posed, at least---not yet.
When what to my horror I suddenly realized:
One present was not wrapped, a present most prized.
For a juggler of the future, three balls all the same.
But how best to foil-wrap, within a tight frame?

Q1. What is the smallest square that can wrap three unit-radius balls, without cutting the square?

To wrap means to completely cover their convex hull. I mention "foil" above because one may want to crinkle the wrapping over sphere caps, analogous to how Mozartkugeln are wrapped.¹

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      [![Balls3][1]][1]

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Likely easier is this question, which may only require rough bounds:

Q2. Which of the two configurations shown above is easier to wrap, easier in the sense that a smaller square suffices?

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¹ The square of diagonal $2 \pi$ is the smallest square that wraps a unit-radius sphere. Demaine, Erik D., Martin L. Demaine, John Iacono, and Stefan Langerman. "Wrapping spheres with flat paper." *Computational Geometry* 42, no. 8 (2009): 748-757. [Journal link](https://www.sciencedirect.com/science/article/pii/S0925772109000182).

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*Added*. I thought I would compute the surface areas of the convex hulls of the two configurations. For the linear configuration,

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          [![Linear3Balls][2]][2]

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I compute $$4 \cdot 2 \pi + 4 \pi = 12 \pi \approx 37.7 \;.$$ For the triangular configuration,

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          [![Tri3Balls][3]][3]

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I compute $4 \pi$ for the three $\frac{1}{3}$ spheres, three times $2 \cdot \pi$ for the cylinder pieces, and two flat $\sqrt{3}$ triangles: $$4 \pi + 6 \pi + 2 \sqrt{3} = 10 \pi + 2 \sqrt{3} \approx 34.9 \;.$$ Of course, this does not address which is easier to wrap with a square.