vector equation is $ x= -3e_2-4 e_3+t(e_1-e_2)$.
I am aware that the equation of a line is in the form of $R=R_0+tv$ but i am confused since my equation of the line seems to be in a different format. This is for study purposes so feel free to use other numbers if it makes it easier to explain.
Hint:
If $\vec e_i$ are the vectors of the standard basis ( as I suppose), than your equation: $$ \vec x= -3 \vec{e_2}-4 \vec{e_3}+t(\vec{e_1}-\vec {e_2}) $$ is the same as: $$ \begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} 0\\-3\\-4 \end{bmatrix} +t\begin{bmatrix} 1\\-1\\0 \end{bmatrix} $$ can you see this?