There are three square shaped plots of land A, B and C, with lengths of each side (in metres) a, b and c, respectively such that a < b < c. The sum of areas of these three plots is 600 m2. There are three rectangular plots of land viz. P (length: b metres, breadth: a metres), Q (length: c metres, breadth: a metres), and R (length: c metres, breadth: b metres). Which of the following cannot be the sum of areas (in square metres) of P, Q and R?

To solve this problem, we first need to express the areas of the square plots A, B, and C, as well as the rectangular plots P, Q, and R, in terms of their side lengths \( a \), \( b \), and \( c \).

### Areas of the Square Plots:
- The area of plot A: \( \text{Area of A} = a^2 \)
- The area of plot B: \( \text{Area of B} = b^2 \)
- The area of plot C: \( \text{Area of C} = c^2 \)

According to the problem, the sum of the areas of the three square plots is 600 square meters:
\[
a^2 + b^2 + c^2 = 600
\]

### Areas of the Rectangular Plots:
- The area of plot P: \( \text{Area of P} = b \times a = ab \)
- The area of plot Q: \( \text{Area of Q} = c \times a = ac \)
- The area of plot R: \( \text{Area of R} = c \times b = bc \)

The sum of the areas of the rectangular plots P, Q, and R is:
\[
\text{Sum of areas of P, Q, and R} = ab + ac + bc
\]

### Expression for the Sum \( ab + ac + bc \):
We know that \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \). We can rearrange this to find \( ab + ac + bc \):
\[
ab + ac + bc = \frac{(a + b + c)^2 - (a^2 + b^2 + c^2)}{2}
\]

Given \( a^2 + b^2 + c^2 = 600 \), the expression simplifies to:
\[
ab + ac + bc = \frac{(a + b + c)^2 - 600}{2}
\]

### Possible Values for the Sum \( ab + ac + bc \):
To determine which value cannot be the sum, consider possible values for \( (a + b + c)^2 \). This value must be greater than 600, since the sum includes additional terms from \( ab + ac + bc \).

The square root of 600 is approximately 24.49, so \( a + b + c \) must be greater than this value. Let's calculate potential \( ab + ac + bc \) values based on reasonable \( a + b + c \) values:

1. \( a + b + c = 25 \): 
   \[
   ab + ac + bc = \frac{625 - 600}{2} = \frac{25}{2} = 12.5 \quad (\text{Not possible since \( ab + ac + bc \) must be an integer.})
   \]
2. \( a + b + c = 26 \): 
   \[
   ab + ac + bc = \frac{676 - 600}{2} = \frac{76}{2} = 38
   \]
3. \( a + b + c = 30 \): 
   \[
   ab + ac + bc = \frac{900 - 600}{2} = \frac{300}{2} = 150
   \]
4. \( a + b + c = 40 \): 
   \[
   ab + ac + bc = \frac{1600 - 600}{2} = \frac{1000}{2} = 500
   \]

From the options considered, the following cannot be a valid sum of areas:
- **12.5** (because the sum must be an integer)
- Any number less than **12.5**

Therefore, **12.5 cannot be the sum** of the areas of P, Q, and R.

### Conclusion:
The possible sums of \( ab + ac + bc \) must be integers and fall within a specific range derived from reasonable values of \( a + b + c \). Among the given options, any non-integer or exceptionally low sum cannot be the sum of areas of P, Q, and R.

Would you like further clarification or details on this solution? Here are some related questions for further exploration:

1. What is the minimum possible value of \( ab + ac + bc \) for the given conditions?
2. How would you solve for \( ab + ac + bc \) if one of the side lengths is known?
3. How can the concept of inequalities help in solving this problem?
4. What would change in the calculations if the sum of areas of A, B, and C were different?
5. Can this problem be solved using a geometric approach? How?

**Tip:** When dealing with quadratic equations and identities, always check if non-realistic solutions (like non-integers in specific contexts) might arise.

Explore how to calculate the areas of squares A, B, and C, and rectangles P, Q, and R. Understand the relationship between the side lengths a, b, and c, and determine which sum cannot be the area of P, Q, and R. Suitable for high school students in Grades 10-12.

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