finding distances between every xyz entry and the last entry [closed] - awk

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I am having some difficulty writing an awk/sed code for finding the distances between every row and the last row systematically. To be more specific, suppose I have a file f1 as follows.
1 2 3
4 5 6
7 8 9
.
.
.
51 52 53
30 31 32
where the first column is the x coordinate, second column is the y coordinate, and third column is the z coordinate. How do I create a file containing the distances between the first row and the last row (i.e. distance between (1,2,3) and (30,31,32)), second row and last row, third row and last row, and so on, until the penultimate row and last row. If f1 has n rows, then the file (let's call it f2) would therefore have n-1 rows.
I have been stuck on this for a long time, but any help would be much appreciated. Thanks!

Answer 1

Use tac to get last line first:
$ tac file | awk '(NR == 1){ x=$1; y=$2; z=$3; next } {
print sqrt((x-$1)^2 + (y-$2)^2 + (z-$3)^2)
}' | tac
50.2295
45.0333
39.8372
36.3731