I am to show that the lines
$$x = y = z$$ and $$x + 1 = \frac{y}{2} = \frac{z}{3}$$
are skew and to find the distance between them.
My attempt: Let $$L_1: x=t \:\:\:\:\:\: y=t \:\:\:\:\:\: z=t$$ and
$$L_2: x=s-1 \:\:\:\:\:\: y = 2s \:\:\:\:\:\: z = 3s.$$
$t=s-1 \Rightarrow s = t+1 \Rightarrow t = 2t+2 \:\: \mbox{and} \:\: t = 3t+3$ which is impossible. So the lines do not have an intersection point and are not parallel because $\langle1,1,1\rangle$ and $\langle1,2,3\rangle$ are not parallel. Therefore the lines are skew.
Then, I have to find the distance between the lines. I sketched the following to do so:
$$\vec{m} = \vec{v_1} \times \vec{v_2} = \langle -1,2,-1 \rangle$$ $$\vec{n} = \vec{m} \times \vec{v_2} = \langle8,2-4\rangle$$
$$\Downarrow$$
$$P: 4x+y-2z+4 = 0$$
$$\Downarrow$$
$$P \cap L_1 : (-\frac{4}{3},-\frac{4}{3},-\frac{4}{3})$$
$$\Downarrow$$
$$L: x = -\frac{4}{3} -t \:\:\:\:\:\: y = -\frac{4}{3} +2t \:\:\:\:\:\: z =-\frac{4}{3} -t$$
$$\Downarrow$$
$L \cap L_2 $ does not exist! But it must be.
Where did I go wrong?