Parallelepiped and shortest path it can take to the opposite vertex

Question

An ant is sitting in a vertex of a right parallelepiped with edges $2, 3, 12$. What is the length of the shortest path it can take to the opposite vertex?

Now, I tried to imagine that and we have a $3D$ box. Just open up the box by unfolding it into a $2D$ figure. now the path from vertex $A$ to vertex $B$ would be a straight line. for $AB$ to be minimum, we need the base to be $(2+2)$ and the width to be $3$.

Answer is $5$. Sounds too easy for me.

Answer 1

Let $ABCDA'B'C'D'$ be our parallelepiped and $AD=12$, $AA'=2$ and $AB=3$. Let our ant be placed in $A$ and wants to go to $C'$.

If he makes it through a point on $A'D'$ or through a point on $BC$, so the shortest part it's $$\sqrt{12^2+(2+3)^2}=13.$$ If he makes it through a point on $DD'$ or through a point on $BB'$, so the shortest part it's $$\sqrt{2^2+(12+3)^2},$$ which is a bit of longer.

Can you end it now?

Answer 2

Here's a pictorial depiction of @Michael Rosenberg's answer. As @Toby Mak commented, you've to find the distance of path from $A$ to $C'$ in the diagram.

There are three potential ways to do so, out of which two are identical. Also, sorry for the missing labels, hope you understand.

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Note that here the height we considered was $3$. If it was $2$, then there would have been one more possible way $$\sqrt{3^2+(12+2)^2}$$ which is obviously not the answer but one of the ways to reach the opposite vertex.