Diwakar starts from Point A and drives 8 km towards West. He then takes a left turn, drives 6 km, turns left and drives 11 km. He then takes a left turn and drives 9 km. He takes a final left turn, drives 3 km and stops at Point P.

How far (shortest distance) and towards which direction should he drive in order to reach Point A again? (All turns are 90 degrees turns only unless specified)

The correct option is 3 km to the south.

Let us trace Diwakar's movement step-by-step using a standard Cartesian coordinate system. Let Point A be the origin (0, 0).
Let North be the positive y-direction, South be the negative y-direction, East be the positive x-direction, and West be the negative x-direction.

1. Starts from Point A (0, 0):
He drives 8 km towards West.
New position = (0 - 8, 0) = (-8, 0).

2. Takes a left turn and drives 6 km:
Since he was driving West, a left turn points him towards South.
New position = (-8, 0 - 6) = (-8, -6).

3. Turns left and drives 11 km:
Facing South, a left turn points him towards East.
New position = (-8 + 11, -6) = (3, -6).

4. Takes a left turn and drives 9 km:
Facing East, a left turn points him towards North.
New position = (3, -6 + 9) = (3, 3).

5. Takes a final left turn, drives 3 km, and stops at Point P:
Facing North, a left turn points him towards West.
Position of Point P = (3 - 3, 3) = (0, 3).

Now, Diwakar is at Point P (0, 3) and wants to reach Point A (0, 0) again.
The difference in coordinates between Point P (0, 3) and Point A (0, 0) is:
x-distance = 0 - 0 = 0 km
y-distance = 0 - 3 = -3 km

Thus, the shortest distance from Point P to Point A is 3 km, and because Point A (0, 0) lies directly below Point P (0, 3) along the y-axis, he must drive towards the South direction.