Determining values of a coefficient for which a system is and isn't consistent. Given the system : \begin{array}{ccccrcc}
x & + & 2y & + & z & = & 3 \\
x & + & 3y & - & z & = & 1 \\
x & + & 2y & + & (a^2-8)z & = & a
\end{array}
Find values of $a$ such that the system has a unique solution, infinitely many solutions, or no solutions.

I begin by placing the system into an augmented matrix.
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
1 & 2 & a^2-8 & a \\
\end{array}
\right]
$
I then perform operations on the matrix.
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
1 & 2 & a^2-8 & a \\
\end{array}
\right]
$ $: (R_3-R_1)\rightarrow$ $\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
0 & 0 & a^2-9 & a-3 \\
\end{array}
\right]
$
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
0 & 0 & a^2-9 & a-3 \\
\end{array}
\right]
$ $: (R_2-R_1; R_3/(a-3))\rightarrow$ $\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & -2 & -2 \\
0 & 0 & a+3 & 1 \\
\end{array}
\right]
$
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & -2 & -2 \\
0 & 0 & a+3 & 1 \\
\end{array}
\right]
$ $: (R_1-2R_2)\rightarrow$ $\displaystyle
\left[
\begin{array}{rrr|r}
1 & 0 & 5 & 3 \\
0 & 1 & -2 & -2 \\
0 & 0 & a+3 & 1 \\
\end{array}
\right]
$

Truth be told, at this point, I'm not too clear on how to progress (or if my approach is even ideal for this issue). I desire to find intervals across all of $a\in\mathbb{R}$ that explain where $a$ causes the system to have a unique, infinite, or inconsistent solution set.
 A: Just from the first reduction you can find:
$$
(a^2-9)z=a-3
$$
Now you simply have to discuss such equation ( the others does not contain the parameter). And you see that this equation has a solution only if $a^2-9 \ne0$ i.e $a \ne \pm 3$.
For $a=3$ the equation becomes: $ 0=0$ that is an identity and this means that the system has infinitely many solutions.
For $a=-3$ the equation becomes $ =-6$ and this means that there are no solutions for the equation and so for the system.
For all other values $a \ne \pm 3$ you find $ z= 1/(a+3)$ and back-substituting in the other equations you find the solutions  $x, y$ of the system that, obviously, depend from the parameter.

Substituting $z=\dfrac{1}{a+3}$ in the first two equations we have:
$$
\begin{cases}
x+2y=3-\dfrac{1}{a+3}\\
x+3y=1+\dfrac{1}{a+3}
\end{cases}
$$
now subtracting the two equations:
$$
y=-2+\dfrac{2}{a+3}=\dfrac{-2(a+2)}{a+3}
$$
and substituting this value in the first equation we have:
$$
x-\dfrac{4(a+2)}{a+3}=3-\dfrac{1}{a+3}
$$
that gives
$$
x=3-\dfrac{1}{a+3}+\dfrac{4a+8}{a+3}=\dfrac{7a+16}{a+3}
$$
A: What strikes me immediately
is that if
$a^2-8 = 1$,
the first and last equations
have the same LHS.
Since choosing $a=3$
makes these equations
identical,
this becomes only two equations
in three unknowns,
so there are an
infinite number of solutions.
On the other hand,
if you choose
$a=-3$,
then the first and third equations
are contradictory,
so there are no solutions
in this case.
I will leave the discussion
of other values of $a$ 
to others.
