Could someone please spot the error I've made in this question:
Q: A theme park has two zip wires. Sarah models the two zip wires as straight lines $(L_1, L_2)$ using coordinates in metres. The ends of one wire are located at $(0,0,0)$ and $(0,100,-20)$. The ends of the other wire are located at $(10,0,20)$ and $(-10,100, -5)$. Use Sarah's model to find the shortest distance between the zip wires:
$$L_1: \vec{r_1}(s) = s(100\hat{j} - 20\hat{k})$$
$$L_2: \vec{r_2}(t) = (10\hat{i}+ 20\hat{k}) + t(-20\hat{i} + 100\hat{j} - 25\hat{k})$$
Let $P$ be a general point on line $L_1$ and $Q$ be a general point on line $L_2$:
Then: $$P = (100s\hat{j} - 20s\hat{k})$$ $$Q = (10-20t)\hat{i} + (100t)\hat{j} + (20 - 25t)\hat{k}$$
Vector $PQ$ is
$$\vec{PQ} = Q - P = ((10-20t)\hat{i} + (100t)\hat{j} + (20 - 25t)\hat{k}) - (100s\hat{j} - 20s\hat{k})$$
$$ \vec{PQ} = (10-20t)\hat{i} + (100t - 100s)\hat{j} + (20 - 25t + 20s)\hat{k}$$
If $\vec{PQ}$ is perpendicular to both lines then: $$\vec{PQ} \cdot \vec{r_1}(s) = \Big((10-20t)\hat{i} + (100t - 100s)\hat{j} + (20 - 25t + 20s)\hat{k}\Big) \cdot (100\hat{j} - 20\hat{k}) = 0$$
$$10000t - 10000s - 400 + 500t - 40s = 0$$ $$10500t - 10040s = 400$$
$$\vec{PQ} \cdot \vec{r_2}(t) = \Big((10-20t)\hat{i} + (100t - 100s)\hat{j} + (20 - 25t + 20s)\hat{k}\Big) \cdot (-20\hat{i} + 100\hat{j} - 25\hat{k}) = 0$$
$$-200 + 400t + 10000t - 10000s - 500 + 625t - 500s = 0$$
$$11025t - 10500s = 700$$
Is this correct? Solving simultaneously gives $t = \dfrac{404}{63}$ and $s = \dfrac{20}{3}$. Plugging this back into vector $\vec{PQ}$ and doing the length of that gives an answer of $121.15$... when I'm fairly certain the answer is $\dfrac{50}{3}$.
I took the liberty of editing beginning of the post so that it was possible to follow your arithmetic. The first set of errors appears in the multiplication marked in red: $20\times 20 = 400$, not $40$. That might have been just a writing error, but in solving for $t$ you should have obtained $t= \frac{44}{63}$ not $\frac{404}{63}$.
I stopped checking after those two errors.