Basis & Dimension for Joint Subspaces Assumption: Assume that $V_1$ and $V_2$ are subspaces of $\mathbb{R}^\mathbb{3}$

Question: "Suppose that $V_1$ is the subspace of $\mathbb{R}^\mathbb{3}$ given by 
  $V_1 = \{(2t-s, 3t, t+2s): t, s \in \mathbb{R}\}$ and 
  $V_2$ is the subspace of $\mathbb{R}^\mathbb{3}$ given by $V_2 = \{(s, t, t): t, s \in \mathbb{R}\}$. Find a basis for $V_1 \cap V_2$ and $dim(V_1 \cap V_2)$."

Where I'm currently at:
I have found the basis for $V_1$ and Basis from $V_2$
which are respectively $\{(2,3,1), (1,0,0)\}$ for $V_1$ and $\{(-1,0,2), (0,1,1)\}$ for $V_2$, and
that $dim(V_1 \cap V_2)$ should be equal or less to $dim(V_1)$ or $dim (V_2)$.
Can you kindly help with the basis of $V_1 \cap V_2$ and $dim(V_1 \cap V_2)$ ?
Any help will be very appreciated
Thank you

The basis for $V_1$ and $V_2$ should be $\{(2,3,1), (-1,0,2)\}, \{(1,0,0), (0,1,1)\}$, respectively. These are two dimensional planes. Their intersection in general should be a $1$-dimensional line.
To find the intersection, you can transform them to equations in terms of $x,y,z$. For $V_1$, we have $x=2t-s, y=3t, z=t+2s$. Some manipulation should give us $z=\frac{5}{3}y-2x$. 
For $V_2$, we see that it is $y=z$. 
Plugging this into the equation for $V_1$, we get the intersection $x=\frac{1}{3}y$. Combining with $y=z$, the basis is $\{(\frac{1}{3}, 1, 1)\}$ since the vectors can be written as $\{(\frac{1}{3}t, t, t)\}$.
