### Triple-Encrypted Vigenere - Please try to crack it

Here's my final take on it. Please take this response with a grain of salt. I am not a professional in the field.

This cipher is seemingly decent on the surface in that since the resulting key is longer than the plain text. However to be secure in the OTP sense, it must be possible to come up with a different set of keys that decode the cipher text into a different human readable message (almost arbitrarily, but limited based off of length)

Example:

Keys "PLM" and "TBEC" with plaintext "HELLOWORLD" (Note that the plain text is longer than the two keys added together)

This encodes to "PQBCSJHEQT". I want to decode this into "PIZZAMYMAN" with some 12 length key starting with "AICDSXJSQG".

The question at hand is whether or not I can generate such a key (Really an arbitrary key) using a key of length 3 and a key of length 4.

This can be simplified as a series of equations

With key "\a \b \c" and "\d \e \f \g" (Slashes signify variables, not letters) We have the key AICDSXJSQG??' split into sections AIC DSX JSQ G??' has relations as

\a = A >> \d = D >> \g = J >> \f = G >> \e

\b = I >> \e = S >> \d = S >> \g = ? >> \f

\c = C >> \f = X >> \e = Q >> \d = ?' >> \g

Or in other words

\d: (\a + \d) % 26 ~ A, (\b + \d) % 26 ~ S, (\c + \d) % 26 ~ Q

\e: (\a + \e) % 26 ~ G, (\b + \e) % 26 ~ I, (\c + \e) % 26 ~ X

\f: (\a + \f) % 26 ~ J, (\b + \f) % 26 ~ ?, (\c + \f) % 26 ~ C

\g: (\a + \g) % 26 ~ D, (\b + \g) % 26 ~ S, (\c + \g) % 26 ~ ?'

This is a fairly straight forward problem now. The one issue now is that we have four equations and three variables. This can lead to an overdetermined system.

Eventually we will have to choose a base for our system, so I'll say that \a = B arbitrarily.

this means that all other pieces are now determined. (Continued)