Linear Equations system The system
$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$
for which $\alpha$ values the linear equation system:


*

*has no solution

*has one solution 

*has more than one solution 


I started to do Gauss elimination on it, but i have no idea what i am looking for and how to approach this, I'm stuck with the Gauss elimination.
My work so far:
\begin{align}
\left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
5 & 2 & 6 & \alpha \\
2 & -1 & \alpha & -6
\end{array}\right)
&\leadsto \left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
0 & 7 & -9 & \alpha + 25 \\
2 & -1 & \alpha & -6
\end{array}\right) \\
&\leadsto \left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
0 & 7 & -9 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right) \\
&\leadsto \left(\begin{array}{rrr|r}
1 & 0 & \alpha + 3 & -1 \\
0 & 7 & -9 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right) \\
\end{align}
 A: Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
5 & 2 & -6 & \alpha \\
2 & -1 & \alpha & -6
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 7 & -21 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right]
&&\begin{aligned} R_2&\gets R_2-5R_1 \\ R_3&\gets R_3-2R_1\end{aligned}
\\[6px]&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 1 & -3 & (\alpha+25)/7 \\
0 & 1 & \alpha-6 & 4
\end{array}\right]
&&R_2\gets \tfrac{1}{7}R_2
\\[6px]&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 1 & -3 & (\alpha+25)/7 \\
0 & 0 & \alpha-3 & (3-\alpha)/7
\end{array}\right]
&&R_3\gets R_3-R_2
\end{align}
If $\alpha\ne3$, the system has unique solution.
If $\alpha=3$, the system has infinitely many solutions.
A: Start with the matrix and data vector
$$
  \mathbf{A} =
\left[
\begin{array}{rrr}
 1 & -1 & 3 \\
 5 & 2 & -6 \\
 2 & -1 & \alpha  \\
\end{array}
\right], \ \alpha \in \mathbb{C}, \quad
%
  b =
\left[
\begin{array}{r}
 -5 \\
 \alpha  \\
 -6
\end{array}
\right]
$$
The reduced row eschelon form is
$$
  \mathbf{E}_{A} =
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right].
$$
The matrix $\mathbf{A}$ has full rank $m=n=\rho=3$. The column vectors span $\mathbb{C}^{3}$.
Select option 2: a solution will always exist, and the solution will be unique. In fact it is
$$
  \mathbf{A} x = b \qquad \Rightarrow \qquad x =
\frac{1}{7}
\left[
\begin{array}{r}
 \alpha - 10 \\
 \alpha + 22  \\
 -1
\end{array}
\right] .
$$
