On a particular day each of Danny, Edwin and Fahim sold three types of pens from their respective shops. As a coincidence, each of them sold an identical number of pens of Type A, an identical number of pens of Type B and an identical number of pens of Type C. However, the numbers of pens of each type that were sold were all different from one another. Also, the three sellers sold each of the types of pens at different prices per unit.

Assertion (A): It is possible that Danny sold each pen of Type A at a profit of Rs 2, each pen of Type B at a loss of Re 1, and each pen of Type C at a loss of Rs 12 and made an overall profit of Rs 8; Edwin sold each pen of Type A at a profit of Rs 2, each pen of Type B at a loss of Rs 6, and each pen of Type C at a loss of Rs 6 and made an overall profit of Rs 14; and Fahim sold each pen of Type A at a profit of Re 1, each pen of Type B at a profit of Re 1, and each pen of Type C at a loss of Rs 3 and made an overall profit of Rs 27.

Reason (R): Framing and solving the three possible linear equations we will find that we get a unique solution.

Given:
Each seller (Danny, Edwin, and Fahim) sells three types of pens (Type A, Type B, Type C) and each sells an identical number of each type of pen.
The profits or losses for each pen type vary among the sellers, and the total profit for each seller is also different.

Assertion (A):
Danny's profit/loss details:
Type A: Profit of ₹2 per pen
Type B: Loss of ₹1 per pen
Type C: Loss of ₹12 per pen
Overall profit = ₹8

So, 

2a -b - 12c = 8 ...(1)

Edwin's profit/loss details:
Type A: Profit of ₹2 per pen
Type B: Loss of ₹6 per pen
Type C: Loss of ₹6 per pen
Overall profit = ₹14

So,

2a -6b -6c = 14..............(2)

Fahim's profit/loss details:
Type A: Profit of ₹1 per pen
Type B: Profit of ₹1 per pen
Type C: Loss of ₹3 per pen
Overall profit = ₹27​​

So, 

a + b -3c = 27..........(3)

Solving the system:

From equations (1) and (2):

Subtracting: (2a - b - 12c) - (2a - 6b - 6c) = 8 - 14

This gives: 5b - 6c = -6 ... (4)

From equation (3): a = 27 - b + 3c ... (5)

Substituting (5) into (1):

2(27 - b + 3c) - b - 12c = 8

54 - 3b - 6c = 8

3b + 6c = 46 ... (6)

Solving equations (4) and (6):

5b - 6c = -6

3b + 6c = 46

Adding: 8b = 40, so b = 5

Substituting back: c = 31/6 ≈ 5.17 and a = 37.5

​Since the number of pens must be whole numbers, this scenario is practically impossible.

​Reason (R): True - The system of three linear equations does indeed have a unique solution when solved algebraically.

So, Assertion (A) is false and Reason (R) is true.

Thus, the correct answer is (d)