$A$ is a $3x3$ matrix. $X$ is the matrix of the unknowns and is $3$ rows by $1$ column. $B$ is the matrix of the solutions and is $3$ rows by a column. And the determinant of $A$ is non-zero ($Det (A) ≠ 0$).
I did the following:
$A.X = B$ Since $Det (A) ≠ 0$, $A$ is reversible. Then, I multiply on both sides of the equality to the left by the inverse of $A$. We obtain: $A^{-1} .A.X = A^{-1}.B$. Then, $X = A^{-1}$. Now, since the inverse of a matrix is unique then the system has a single solution. Then, there are no scalars $a$, $b$, $c$ such that $A.X = B$ has infinite solutions and neither are such that the system has no solutions. The scalars are the elements that form the matrix of solutions
Are my demonstration and justification correct?
You did not explain what these $a,b,c$ really are. But what you're trying to write is correct, excepting that $X=A^{-1}B$, and not $A^{-1}$.
I'm supposing that you've tried to say that $B=[a\,\,b\,\,c]^{t}$. Even that way the system have one, and only one solution, because you're supposing $det(A)\neq0$. Doesn't matter what B is. Further, you can solve the system $AX=B$ in function of $a,b,c$, and find the solution $S$ for whatever you put into matrix $B$.