What is the concept behind Miller Indices? Have a look at the attached image below.
If I look at the image intuitively, my common sense tells me that the Miller Indices for the vector should be [1,1,-2,1] since we're travelling one unit along a1 vector, one unit along a2 vector, one unit along z vector and due to vector addition, negative one unit along a3 direction.
However, using the formulae for conversion of the Hexagonal crystal to a Cubic crystal and then back to the hexagonal crystal, the Miller Indices turn out to be [1,1,-2,3] !!
For those interested in the details, the video can be found here.
So what exactly is the basis behind the concept being used here? Why does it seem so non-intuitive to me?

 A: I think I understand what you have done and the question that you have asked about the result that you have obtained.  
You started with [1 1 -2 1] in 4-index notation and then transformed it into [1 1 1] in 3-index notation.  
Then you started with [1 1 1] and attempted to convert it back into 4-indev notation.
From the first two indices of the 3-index notation you obtained $+\frac 13$, $+\frac 13$ and $-\frac 23$ as the first three indices in the 4-index notation.  
Now how to convert from the third index of the 3-index notation to the 4-index notation?
What you did was to say that the last index in the 3-index notation which is a one transforms to become a one in the 4-index notation?
You multiplied by 3 to get only integers which resulted in a 4-index result of [1 1 -2 3] which has mystified you.  
The error that you have made is to say that there is no change in the last index.
You transformation of [1 1 .] yielded [$\frac 13$ $\frac 13$ -$\frac 2 3$  .]
Now [$\frac 13$ $\frac 13$ -$\frac 2 3$  .] is the projection of your vector onto the $z=0$ plane and is only $\frac 13^{\rm rd}$ of the projection needed to reach the edge of the unit cell.  
So the height of the vector above the $z=0$ plane is $\frac 13$ of the height of the unit cell.  
The correct transformation of [1 1 1] is [$\frac 13$ $\frac 13$ -$\frac 2 3$ $\frac 1 3$] which leads to the result that you were expecting.
