How to think about the following problem.
Let $R$ be a Euclidean domain generated by $a,b,$ and $c$. Then does there exists an invertible $3\times3$ matrix such that $a,b,$ and $c$ appear in one row?
I am trying to solve by taking $R$ as a module over itself and trying to find a module homomorphism taking $a$ to $a^2$, $b $ to $b^2$ and $c$ to $c^2$ but not able to succeed.
Thank you.
Since the OP wants $a,b,c,$ to appear in a row of an invertible matrix, I am guessing that the assumption means that the ideal (aka $R$-submodule) of $R$ generated by $a,b,c$ is $R$, that is $Ra+Rb+rc=R$
But we hav the following general result.
Let $R$ be a PID (eg, $R$ is an Euclidean domain), let $M$ be a free $R$-module of finite rank, and let $N$ be a submodule of $M$. Let $a_1,\ldots,a_r\in R$ be the invariant factors of $N$ wrt to $M$. Then $N$ is a direct factor of $M$ if and only if $a_1,\ldots,a_r$ are all units.
Put in other words: let $(e_1,...,e_n)$ and $a_1,...,a_r\in R$ such that $(e_1,...,e_n)$ is a basis of $M$ and $(a_1e_1,..,a_re_r)$ is a basis of $N$.
Then there exists a submodule $Q$ of $M$ such that $M=N\oplus Q$ if and only if $a_1,...,a_r$ are all units of $R$.
This should be in any good introductory book on modules.
Now, if you take $M=R^n$ and $v\in R^n$ nonzero, then the unique invariant factor of $N=Rv$ is just a gcd of the coordinates of $v$.
If you translate the previous result, you will get that $v$ may be extended to a basis of $R^n$ if and only if the ideal (aka $R$-submodule) generated by the coordinates of $v$ is $R$, which is exactly what you are looking for (after an eventual transposition).