Manish starts from Point A and drives $12\text{km}$ towards East. He then takes a left turn, drives $5\text{km}$, turns left and drives $30\text{km}$. He then takes a left turn and drives $20\text{km}$. He takes a final left turn, drives $18\text{km}$ and stops at Point Q. How far (shortest distance) and towards which direction should he now drive to reach Point A again? (All turns are $90^{\circ }$ turns only.)

Let's map Manish's movement step-by-step using Cartesian coordinates, assuming Point A is at the origin $(0,0)$:
1. **Starts at A $(0,0)$ **, drives $12\text{km}$ East $\to$ reaches $(12,0)$.
2. Takes a left turn (faces North), drives $5\text{km}$ $\to$ reaches $(12,5)$.
3. Turns left (faces West), drives $30\text{km}$ $\to$ new X-coordinate $=12-30=-18$. Reaches $(-18,5)$.
4. Takes a left turn (faces South), drives $20\text{km}$ $\to$ new Y-coordinate $=5-20=-15$. Reaches $(-18,-15)$.
5. Takes a final left turn (faces East), drives $18\text{km}$ $\to$ new X-coordinate $=-18+18=0$. Reaches Point Q at $(0,-15)$.

Now, Manish is at Point Q $(0,-15)$, and he wants to return to Point A $(0,0)$.

* The X-coordinates are the same ($0$).
* The vertical distance is $0-(-15)=15\text{km}$.
* Since Point A is directly above Point Q along the Y-axis, he must drive North to reach Point A.

Therefore, he should drive $15\text{km}$ towards North.