`|{:(,"List-I(Atom/Ions)",,"List-II(Electron Affinity in eV/atom and consider that EA"=-Delta H_(e.g)")"),((P),F,(1),3.4),((Q),F^(+),(2),17.4),((R ),Cl,(3),13),((S),Cl^(+),(4),3.6):}|`
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Options

A. (A+B+C) is an integer
B. (A+B+C) is an integer
C. (8A+2C)is an integer
D. A,BC are integers

Answer

To solve the equation \( 2a^2 + 3b^2 = 35 \) where \( a \) and \( b \) are integers, we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 2a^2 + 3b^2 = 35 \]
We will isolate \( b^2 \):
\[ 3b^2 = 35 - 2a^2 \]
\[ b^2 = \frac{35 - 2a^2}{3} \]

### Step 2: Finding Integer Solutions
For \( b^2 \) to be an integer, \( 35 - 2a^2 \) must be divisible by 3. We will check values of \( a \) that keep \( b^2 \) non-negative.

### Step 3: Testing Integer Values for \( a \)
We will test integer values for \( a \):

1. **For \( a = 0 \)**:
\[ b^2 = \frac{35 - 2(0)^2}{3} = \frac{35}{3} \quad \text{(not an integer)} \]

2. **For \( a = 1 \)**:
\[ b^2 = \frac{35 - 2(1)^2}{3} = \frac{33}{3} = 11 \quad (b = \pm \sqrt{11} \text{ (not an integer)}) \]

3. **For \( a = 2 \)**:
\[ b^2 = \frac{35 - 2(2)^2}{3} = \frac{27}{3} = 9 \quad (b = \pm 3) \]

4. **For \( a = 3 \)**:
\[ b^2 = \frac{35 - 2(3)^2}{3} = \frac{23}{3} \quad \text{(not an integer)} \]

5. **For \( a = 4 \)**:
\[ b^2 = \frac{35 - 2(4)^2}{3} = \frac{19}{3} \quad \text{(not an integer)} \]

6. **For \( a = 5 \)**:
\[ b^2 = \frac{35 - 2(5)^2}{3} = \frac{5}{3} \quad \text{(not an integer)} \]

7. **For \( a = 6 \)**:
\[ b^2 = \frac{35 - 2(6)^2}{3} = \frac{-7}{3} \quad \text{(not valid)} \]

### Step 4: Collecting Solutions
From our tests, we found valid pairs:
- \( (2, 3) \)
- \( (2, -3) \)
- \( (-2, 3) \)
- \( (-2, -3) \)

### Step 5: Additional Valid Pairs
We can also check negative values for \( a \):
- **For \( a = -2 \)**, we get the same results as for \( a = 2 \).

### Final Count of Solutions
The valid integer pairs \( (a, b) \) are:
1. \( (2, 3) \)
2. \( (2, -3) \)
3. \( (-2, 3) \)
4. \( (-2, -3) \)
5. \( (4, 1) \)
6. \( (4, -1) \)
7. \( (-4, 1) \)
8. \( (-4, -1) \)

Thus, we have a total of **8 distinct pairs**.

### Conclusion
The total number of integer solutions is:
\[ \text{Answer: } 8 \]