Hi Simon

I think you'll find that the if statement is misguided, since it treats
numbers as decimal rather than hex valued.

Happily, this simplifies the solution.

#lang racket

(define-syntax-rule (hex h ...)
  (bytes (string->number (~a (quote h)) 16) ...))

(hex a b c 1 2 3 41 42 43) ; #"\n\v\f\1\2\3ABC"


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