I want to find the intersection vectors between 2 spans, something like this: $span\left \{ \begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix},\begin{bmatrix} 1\\ 2\\ 2 \end{bmatrix} \right \}\cap span\left \{ \begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix},\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix} \right \}$ Visually, I know their intersection is going to be a line. But I have to find the vector of this line. I attempt to put them into an equation and solve for it this way: $\begin{align*} a\begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix}+ b\begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} = c\begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix}+ d\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix} \\ a\begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix}+ b\begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} - c\begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix}- d\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix}=0 \\ \begin{bmatrix} 3 & 1 & -3 & -8\\ 6 & 2 & 0 & -5\\ 0 & 2 & 1& 6 \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}=0 \\ \Rightarrow \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}=t\begin{bmatrix} \frac{5}{12}\\ -\frac{25}{12}\\ -\frac{11}{6}\\ 1 \end{bmatrix}, t\in \mathbb{R}\end{align*} $
So now, I assume I could plug in those values back into the first equation: $\frac{5}{12}\begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix}+ -\frac{25}{12}\begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} = -\frac{11}{6}\begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix}+ 1\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix}$
And claim that the vector that this equation is equals to each other is the vector span of the intersection. But when I try to add them together, I get this: $\begin{bmatrix} -\frac{5}{6}\\ -\frac{5}{3}\\ -\frac{25}{6} \end{bmatrix} \neq \begin{bmatrix} \frac{5}{2}\\ 5\\ -\frac{25}{6} \end{bmatrix}$ They are not equal! That's weird.
How come they don't equal to each other? Is what I am doing right? How and what other ways can I use to find the intersection between the 2 vector spans?