For which values of a do the following vectors for a linearly dependent set in $R^3$? For which values of a do the following vectors for a linearly dependent set in $R^3$?
$$V_1= \left(a,\, \frac{-1}{2}, \,\frac{-1}{2}\right),\;\; V_2= \left(\frac{-1}{2},\, a, \,\frac{-1}{2}\right),\; \;V_3= \left(\frac{-1}{2}, \,\frac{-1}{2},\, a\right)$$
Please would it be possible to advise me how I would go about solving this?
I'm not sure if I should be using only row reduction because I think it may relate to eigenvalues and eigenvectors but we haven't covered those concepts in class as yet.
Am I supposed to find the determinant?
 A: We want to find all (only) those value(s) that will make the vectors linearly dependent. 
Can you see, for example,  why $\,a = -\frac 12\,$ is a problem? Why would $\bf \,a = -\frac 12\,$ make the vectors linearly dependent? And why would $\bf\, a = 1\,$ make the vectors linearly dependent?
A set of vectors, in your case, in $\mathbb R^3$, is linearly dependent if any one of them can be written as a linear combination of the others. In either of the above cases, $\,a = -\frac 12, \,\text{ or}\; a = 1,\,$ one or more of the vectors can be expressed as a linear combination of the others.

Recall: The determinant of an $n\times n$ matrix equals zero $\iff$ (when and only when) its column vectors are linearly dependent. 
So you can also solve for $a$ by setting up the matrix using your vectors as columns in a $3 \times 3$ matrix; find the determinant, which will be a function of $a$, set it equal to zero, and solve for the $a$ values that make the determinant equal to zero (find the zeros of the determinant). Those and only those values are values for which the vectors are linearly dependent. 
A: Row reduction has little to do with eigenvalues, but it has much to do with linear dependence. Actually it's the method of choice.
It's better to change the order into $v_3,v_2,v_1$
$$
\begin{bmatrix}
-\frac{1}{2} & -\frac{1}{2} & a \\
-\frac{1}{2} & a & -\frac{1}{2} \\
a & -\frac{1}{2} & -\frac{1}{2}
\end{bmatrix}\to
\begin{bmatrix}
1 & 1 & -2a \\
1 & -2a & 1 \\
-2a & 1 & 1
\end{bmatrix}\to
\begin{bmatrix}
1 & 1 & -2a \\
0 & -2a-1 & 1+2a \\
0 & 1+2a & 1-4a^2
\end{bmatrix}
$$
If $a=-1/2$ the matrix becomes
$$
\begin{bmatrix}
1 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
$$
Otherwise we can continue the reduction
$$
\to
\begin{bmatrix}
1 & 1 & -2a \\
0 & 1 & -1 \\
0 & 0 & 2 + 2a - 4a^2
\end{bmatrix}
$$
The last row is zero (and the vectors are linearly dependent) if and only if
$$
2a^2 - a - 1 = 0
$$
that is
$$
a=\frac{1\pm\sqrt{1+8}}{4}=\frac{1\pm3}{4}
$$
which gives $a=-1/2$ (not good now as we were assuming $a\ne-1/2$) or $a=1$.
Conclusion: the three vectors are linearly dependent if and only if $a=-1/2$ or $a=1$.

For $a=-1/2$, the reduced form says that the general form for getting
$$
\alpha_3v_3 + \alpha_2v_2 +\alpha_1 v_1=0
$$
is to set arbitrary values for $\alpha_2$ and $\alpha_1$ and taking $\alpha_3=-\alpha_2-\alpha_1$. For instance, with $\alpha_1=1$ and $\alpha_2=0$ we get
$$
v_1-v_3=0.
$$
For $a=1$, the reduced form is
$$
\begin{bmatrix}
1 & 1 & -2 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}
$$
that can be further reduced to row echelon form
$$
\begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}
$$
which says that the general form for getting
$$
\alpha_3v_3 + \alpha_2v_2 +\alpha_1 v_1=0
$$
is to set an arbitrary value for $\alpha_1$ and setting $\alpha_3=\alpha_1$, $\alpha_2=\alpha_1$; for instance, with $\alpha_1=1$, we have
$$
v_1+v_2+v_3=0.
$$
The order of the vectors is irrelevant, because addition of vectors is commutative.
A: To be linearly dependent in $\mathbb{R}^3$, the three vectors would have to be co-planar.  One test would be that the triple product of the three vectors (in any order) would be zero (since a vector perpendicular to the mutual perpendicular of any two others would have to be in the same plane as those two).  In Cartesian components, the triple product can be expressed as a 3 x 3 determinant of the three vectors.  So you are looking for values of $a$ that make that determinant (using your vectors as either rows or columns) equal to zero.
[Note:  $a = -1/2$ is the "trivial" solution; there is another value that works...]
