3D geometry - equation of line Question:

Suppose we have the two lines: $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ and $$\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$
  Find the equation of the line which covers the shortest distance between the two given lines.

Attempt:
I know that the line that we have to find will be along the cross product of the direction ratios of the two given lines (considering equation of line in form $\vec a + t\vec b$). But how will I find $\vec a$?
 A: The lines can be written in parametric form as $$\underline{r}=\left (\begin{matrix}1+2t\\2+3t\\3+4t\end{matrix}\right)$$ and  $$\underline{r}=\left (\begin{matrix}2+3s\\4+4s\\5+5s\end{matrix}\right)$$
Subtracting these gives us a displacement vector which joins two arbitrary points, one on each line. This vector must be perpendicular to the direction vectors of the given lines, so we can form and solve a simultaneous equation for $t$ and $s$.
 $$\left (\begin{matrix}-1+2t-3s\\-2+3t-4s\\-2+4t-5s\end{matrix}\right)\cdot\left (\begin{matrix}2\\3\\4\end{matrix}\right)=0$$
$$\Rightarrow29t-38s=16$$ and $$\left (\begin{matrix}-1+2t-3s\\-2+3t-4s\\-2+4t-5s\end{matrix}\right)\cdot\left (\begin{matrix}3\\4\\5\end{matrix}\right)=0$$
$$\Rightarrow38t-50s=21$$
Solving these, we get $$s=-\frac 16, t=\frac 13$$
Using these values we can get the direction of the line as $$\left(\begin{matrix}\frac 16\\-\frac 13\\\frac 16\end{matrix}\right)\text{ or }\left(\begin{matrix}1\\-2\\1\end{matrix}\right)$$
and using the value of $t$ in the first equation, we have a fixed point $$\left(\begin{matrix}\frac 53\\3\\\frac{13}{3}\end{matrix}\right)$$
Hence the equation of the line can be written as $$\underline{r}=\left(\begin{matrix}\frac 53\\3\\\frac{13}{3}\end{matrix}\right)+\lambda\left(\begin{matrix}1\\-2\\1\end{matrix}\right)$$
A: The line should be intersect with both given line.
Suppose the 2 intersections are $P=\vec c+u\vec d$ on the $1^{st}$ line and $Q=\vec e+v\vec f$ on the second line, then $$\begin{align}\vec{P}-\vec{Q}=(\vec c+u\vec d)-(\vec e+v\vec f)=&t\vec{b}\\t\vec{b}-u\vec{d}+v\vec{f}=&\vec{c}-\vec{e}\\\begin{bmatrix}\vec{b}&\vec{d}&\vec{f}\end{bmatrix}*\begin{bmatrix}t\\u\\v\end{bmatrix}=&\begin{bmatrix}\vec{c}-\vec{e}\end{bmatrix}\end{align}$$
So $$\begin{bmatrix}t\\u\\v\end{bmatrix}=\begin{bmatrix}\vec{b}&\vec{d}&\vec{f}\end{bmatrix}^{-1}*\begin{bmatrix}\vec{c}-\vec{e}\end{bmatrix}$$
Solving (t,u,v), it is sufficient to let $$\vec{a}=\vec{P}$$
