For any square linear system $\,A\vec x=\vec b\,$ over some field, there exists a unique solution iff $\,\det A\neq 0\,$ , as then we can use the inverse matrix:
$$A\vec x=\vec b\Longleftrightarrow A^{-1}A\vec x=A^{-1}\vec b\Longleftrightarrow A^{-1}\vec b=\vec x $$
As for (a) and your "main question": if $\,\det A=0\,$ one still may have to check whether there are no solutions or infinite solutions (assuming we're working on an infinite field). For example, if the system is homogeneous (over an infinite field) it must have infinite solutions, whereas if the system is non-homogeneous it may have no solutions or several:
$$\begin{cases}x+y=1\\x+y=1\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{1}\longrightarrow\,\,\text{infinite solutions}$$
$$\begin{cases}x+y=1\\x+y=0\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{0}\longrightarrow\,\,\text{no solutions at all}$$
and, of course, in both cases above we have $\,\det A=0\,$
$$
\begin{pmatrix}
2 & -1 & 3 & 5\\
4 & 2 & 2 & 6\\
-2 & \alpha & 3 & 1
\end{pmatrix}\longrightarrow
\begin{pmatrix}
2 & -1 & 3 & 5\\
0 & 4 & \!\!\!-4 & \!\!\!-4\\
0 & \alpha-1 & 6 & 6
\end{pmatrix}
$$
So how can you do to have one row linearly dependent on the other two?
Spoiler!
$\,\alpha=-5\,$
Best Answer
One can write your system $A x = b$ as augmented matrix and bring it into row echelon form $$ \left[ \begin{array}{rrr|r} 2 & 4 & -2 & 0 \\ 3 & 5 & 0 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 3 & 5 & 0 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 0 & -1 & 3 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 0 & 1 & -3 & -1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 5 & 2 \\ 0 & 1 & -3 & -1 \end{array} \right] $$ This translates back into $$ x + 5z = 2 \\ y - 3z = -1 $$ or $x = (2-5z, -1+3z, z)^T$, where $z \in \mathbb{R}$. So there are infinite many solutions.
From a geometric point of view, each equation defines an affine plane in $\mathbb{R}^3$, which is a plane, not necessarily including the origin. $$ (2, 4, -2) \cdot (x,y,z) = 0 \\ (3, 5, 0) \cdot (x, y, z) = 1 $$ The first plane has a normal vector $(2,4,-2)^T$ and includes the origin, the second plane has normal vector $(3,5,0)$ and is $1/\sqrt{3^2 + 5^2}$ away from the origin.
The solution of the system is the intersection of those two planes. And only the cases empty intersection, intersection is a line or intersection is a plane can happen. Here the intersection is a line.
The image shows the two intersecting planes, the intersection line, and the point $P$ which corresponds to the solution with $z = 0$.