Finding a basis for the intersection of two vector subspaces. 
Suppose:
$V_1$ is the subspace of $\mathbb{R}^3$ given by $V_1 =
 \{(2t-s,t,t+s)|t,s\in\mathbb{R}\}$
and
$V_2$ is the subspace of $\mathbb{R}^3$ given by $V_2 =
 \{(s,t,s)|t,s\in\mathbb{R}\}$.

How could a basis and dimension for $V_1 \cap V_2$ be determined?
I thought of maybe setting the respective components of the two spaces equal to each other such that $2t-s = s$, $t=t$, and $t+s=s$.
Solving each of these equations it can be found that $2t=2s$, $t=t$, and $t=0 \implies s=0 \implies V_1 \cap V_2 = \{(0,0,0)\} \implies \text{dim}(V_1 \cap V_2)=0$.
This seems like it isn't right though, but I'm really uncertain.

In general, to find the intersection of two subspaces, you express a vector in the intersection in two different ways and then solve the system of linear equations that result, a la copper.hat’s answer.  
In the special case of $\mathbb R^3$, there’s a quick and dirty way to find the intersection of two different planes through the origin. Since their intersection will be orthogonal to both normals, you can get the answer by computing three cross products—two to compute the normals and another to find a vector orthogonal to them. With only 3 dimensions, this last vector must lie on the line of intersection.  
To illustrate, the plane $V_1$ is spanned by $(-1,0,1)^T$ and $(2,1,1)^T$, so a normal is $(-1,0,1)^T\times(2,1,1)^T=(-1,3,-1)^T$, while $V_2$ is spanned by $(1,0,1)^T$ and $(0,1,0)^T$ with normal $(1,0,1)^T\times(0,1,0)^T=(-1,0,1)^T$. The cross product of these normals is $(-1,3,-1)^T\times(-1,0,1)^T=(3,2,3)^T$, just as was found by the general method.  
This also gives a way to find the intersection of two lines in the plane or the line through two points in the plane: use homogeneous coordinates and take their cross product.
