First of all, before any deeper looking in the question, it is easy to see that for k=0, the last two vectors are obviously dependent (since they are the same), and for k=1 the first two vectors are dependent (since they are the same).
But, essentially we want to find when there are a,b,c s.t. a*v1+b*v2+c*v3=0, and at least one of the variables isn't zero.
$$a(k,1,1)+b(1,1,k)+c(1,1,0)=0$$
$$(ka,a,a)+(b,b,kb)+(c,c,0)=0$$
$$(ka+b+c,a+b+c,a+kb)=(0,0,0)$$
thus:
ka+b+c = 0
a+b+c = 0
a+kb = 0
We now want to solve these equations, so we'll put them in a matrix, and see when the rows in the rows in the matrix are dependent, by changing it to the canonical form:
$$\begin{bmatrix}k & 1 & 1 \\ 1 & 1 & 1 \\ 1 & k & 0 \end{bmatrix}$$
If we get the diagonal I canonical form, the only answer is the trivial answer (a=b=c=0), and the vectors aren't dependent. if not - there is set of scalars a,b,c that shows the vectors are dependent.
Notice that the columns of the matrix are actually the vectors themselves. As you can see, this should happen no matter what were the vectors to begin with, so you can simply skip to the matrix above instead of going through the equations every time. Also, since the rows o a matrix are dependent iff the columns are dependent, you can also use the vectors as rows instead of columns, if you prefer.
after playing with the we matrix we will get:
$$\begin{bmatrix}1 & 1 & 1 \\ 0 & k-1 & -1 \\ 0 & 0 & k \end{bmatrix}$$
if k=0 the rows are dependent. if not then we can divide the third row by k, and afterwards subsract the third row from the other two rows, getting:
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & k-1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
if k=1, again we'll get a row of zero's (the second row), but if not, we can divide the second row in k-1, and then get $I_3$, meaning the rows of the original matrix were not dependent.
Now, if we want to find the combinations themselves, we only need to examine the canonical forms of the matrix, and solve the equations it represents. For example, for k=0, we have:
$$\begin{bmatrix}1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{bmatrix}=>\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$
$$b+c=0$$ $$ a=0$$
If we choose c=1, we get b=-1,a=0. and that's one linear combination that proves the vectors are dependent when k=0.
for k=1, the same process will lead us to a=1,b=-1,c=0.
In conclusion, the method above shows that to find if vectors are dependent we can simply put them in a matrix, and see if we get the I canonical form (meaning they are not dependent), or a canonical form with a row of zero's (meaning they are). In this case, they are dependent iff k=0,1.
A more explicit way to do this would be by considering
\begin{align} v_1 = (\lambda,a,a),\\ v_2 = (a,\lambda,a), \\ v_3 = (a,a,\lambda),\end{align}
such that $a\in\mathbb{R}$ is fixed. (In this case, $a=-1/2$.)
We want to find $\lambda$ such that $v_1,v_2,v_3$ are LD. If these vectors are LD, then $\exists c,d \in\mathbb{R}: cv_1+dv_2=v_3$. That is:
\begin{align}\begin{cases}c\lambda+da=a \\ ca+d\lambda=a \\ (c+d)a=\lambda\end{cases}\implies \begin{cases}(c+d)(a+\lambda) = 2a \\ (c+d)a=\lambda\end{cases}\implies \begin{cases}(c+d+1)\lambda = 2a \\ (c+d)a=\lambda\end{cases}\end{align}
And we can deduce for $a,\lambda\neq 0$:
\begin{align}&(c+d)(c+d+1)\lambda a = 2\lambda a \\ \implies &(c+d)(c+d+1)= 2 \\ \implies &(c+d) = 1 \text{ or } -2.\end{align}
So, from $(c+d)a=\lambda$, we have $\lambda = -2a,\lambda = a$. In the case $a=-1/2$ we have, indeed, $\lambda=1,\lambda=-1/2$.
Best Answer
Yes. But in general you will not end up with a square matrix, so the appropriate concept is the rank of the matrix (i.e. the number of leading ones in the row-reduced form); for your vectors to be linearly independent it has to agree with the number of vectors.