How to eliminate the list case by `Eliminate` function

It is known that line L1: $\frac{x-a_{2}}{a_{1}}=\frac{y-b_{2}}{b_{1}}=\frac{z-c_{2}}{c_{1}}$ and line L2: $\frac{x-a_{3}}{a_{2}}=\frac{y-b_{3}}{b_{2}}=\frac{z-c_{3}}{c_{2}}$ intersect at a point.

Now I want to find the possible relationship between vectors $a_{1}=\left[\begin{array}{l} a_{1} \\ b_{1} \\ c_{1} \end{array}\right]$, $a_{2}=\left[\begin{array}{l} a_{2} \\ b_{2} \\ c_{2} \end{array}\right]$, $a_{3}=\left[\begin{array}{l} a_{3} \\ b_{3} \\ c_{3} \end{array}\right]$.

But I had trouble solving this problem with the following code:

Eliminate[(x0 - a2)/a1 == (y0 - b2)/b1 == (z0 - c2)/c1 && (x0 - a3)/ a2 == (y0 - b3)/b2 == (z0 - c3)/c2 && α1 = {a1, a2, a3} && α1 == {a1, a2, a3} && α2 == {b1, b2, b3} && α3 == {c1, c2, c3}, {x0, y0, z0}](*Suppose that the intersection of two lines is {x0, y0, z0}*) 

Even I can't get the relationship between vectors with specific values:

Solve[α1 == {1, 2, 3} && α2 == {1, 0, 0} && α3 == {2, 2, 3} && α3 == k1*α1 + k2*α2, {k1, k2}, {α1, α2, α3}] Reduce[α1 == {1, 2, 3} && α2 == {1, 0, 0} && α3 == {2, 2, 3}, {α1, α2, α3}](*we should get a result like α3 == α2 + α1*) 

Eliminate function seems to be unable to directly deal with the relationship in the form of vector, what should I do to better solve this problem?