You say "But we've only learned to gleam whether or not a set is dependent/independent" ("gleam" is not the right word here!) so use that.
You don't say whether the 3 by 1 vectors are \(\displaystyle \begin{bmatrix}1 & 1 & 1\end{bmatrix}\), \(\displaystyle \begin{bmatrix}0 & 1 & 2\end{bmatrix}\), \(\displaystyle \begin{bmatrix}1 & 2 & k \end{bmatrix}\) or \(\displaystyle \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}\), \(\displaystyle \begin{bmatrix}1 \\ 1 \\ 2 \end{bmatrix}\), \(\displaystyle \begin{bmatrix}1 \\ 2 &\\ k \end{bmatrix}\) but it really doesn't matter.
A set of three vectors, u, v, w, is "dependent" if there exist numbers, a, b, c, not all zero, such that au+ bv+ cw= 0. The vectors are "independent" if that is not true.
So if your vectors are the first set, you want to look at \(\displaystyle a\begin{bmatrix}1 & 1 & 1 \end{bmatrix}+ b\begin{bmatrix}0 & 1 & 2 \end{bmatrix}+ c\begin{bmatrix}1 & 2 & k\end{bmatrix}\)\(\displaystyle = \begin{bmatrix}a+ c & a+ b+ 2c & a+ 2b+ kc\end{bmatrix}\)\(\displaystyle = \begin{bmatrix}0 & 0 & 0 \end{bmatrix}\). That is the same as the three equations a+ c= 0, a+ b+ 2c= 0, and a+ 2b+ kc= 0. If you subtract the first equation from the second, a cancels and you get b+ c= 0. If you subtract the first equation from the third, a also cancels and you get 2b+ (k-1)c= 0. If you subtract the second of those two equations from twice the first b cancels leaving 2c- (k-1)c= (3- k)c= 0. For all except one value of k, that gives c= 0 so the vectors are independent. For what value of k is c not necessarily 0 so the vectors are dependent?
For the second set, do the same thing. You want to look at \(\displaystyle a\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}+ b\begin{bmatrix}1 \\ 1 \\ 2 \end{bmatrix}+ c\begin{bmatrix}1 \\ 2 \\ k\end{bmatrix}= \begin{bmatrix} a+ b+ c \\ b+ 2c \\ a+ 2b+ kc \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\). That is the same as the three equations a+ b+ c= 0, b+ 2c= 0, and a+ 2b+ kx= 0. If you subtract the first equation from the third a cancels and you have b+ (k-1)c= 0. If you subtract the second equation from that b cancels and you have (k- 1)c- 2c= (k- 3)c= 0. For all except one value of k, that gives c= 0 so the vectors are independent. For what value k is c not necessarily 0 so the vectors are dependent.
Of course, the simpler way to handle this is to argue that the set of vectors is independent if and only if the matrix formed of these vectors is invertible so you could, as staple suggests, determine whether your give matrix is invertible by calculating the determinant and setting it equal to 0. That will be an equation in k so solve the equation for k. Probably the simplest way would be to row-reduce the matrix and determine what value of k will give a row all zero.
