`x#y={{:(x+2y,if,|x+y|" is even"),(2x" @2 "y,if,|x+y|" is odd"):}`
`x"@"y={{:(3x-y,if,|x+y|" is even"),(2x#2y" ,if",|x+y|" is odd"):}`
Which of the following is true ?
(i) `6"@"2=12#2`
(ii) `6#4=(14"@"2)+(2"@"5)`
(iii) `(6"@"2)+(6#3)=(12#2)+(14"@"12)-(2"@"6)`

Options

A. (i) and (ii) only
B. (i) and (iii) only
C. (ii) and (iii) only
D. (i), (ii) and (iii)

Answer

To solve the problem, we need to evaluate the three statements given using the definitions of the operations \( x \# y \) and \( x @ y \).

### Definitions:
1. \( x \# y = \begin{cases} x + 2y & \text{if } |x+y| \text{ is even} \\ 2x @ 2y & \text{if } |x+y| \text{ is odd} \end{cases} \)

2. \( x @ y = \begin{cases} 3x - y & \text{if } |x+y| \text{ is even} \\ 2x \# 2y & \text{if } |x+y| \text{ is odd} \end{cases} \)

### Evaluating the Statements:

#### Statement (i): \( 6 @ 2 = 12 \# 2 \)

1. **Calculate \( 6 @ 2 \)**:
- \( |6 + 2| = 8 \) (even)
- Use the first case of \( @ \):
\[ 6 @ 2 = 3(6) - 2 = 18 - 2 = 16 \]

2. **Calculate \( 12 \# 2 \)**:
- \( |12 + 2| = 14 \) (even)
- Use the first case of \( \# \):
\[ 12 \# 2 = 12 + 2(2) = 12 + 4 = 16 \]

3. **Conclusion for Statement (i)**:
\[ 6 @ 2 = 16 \quad \text{and} \quad 12 \# 2 = 16 \quad \Rightarrow \quad 6 @ 2 = 12 \# 2 \quad \text{(True)} \]

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#### Statement (ii): \( 6 \# 4 = (14 @ 2) + (2 @ 5) \)

1. **Calculate \( 6 \# 4 \)**:
- \( |6 + 4| = 10 \) (even)
- Use the first case of \( \# \):
\[ 6 \# 4 = 6 + 2(4) = 6 + 8 = 14 \]

2. **Calculate \( 14 @ 2 \)**:
- \( |14 + 2| = 16 \) (even)
- Use the first case of \( @ \):
\[ 14 @ 2 = 3(14) - 2 = 42 - 2 = 40 \]

3. **Calculate \( 2 @ 5 \)**:
- \( |2 + 5| = 7 \) (odd)
- Use the second case of \( @ \):
\[ 2 @ 5 = 2(2) \# 2(5) = 4 \# 10 \]
- Now calculate \( 4 \# 10 \):
- \( |4 + 10| = 14 \) (even)
- Use the first case of \( \# \):
\[ 4 \# 10 = 4 + 2(10) = 4 + 20 = 24 \]

4. **Combine Results**:
\[ (14 @ 2) + (2 @ 5) = 40 + 24 = 64 \]

5. **Conclusion for Statement (ii)**:
\[ 6 \# 4 = 14 \quad \text{and} \quad (14 @ 2) + (2 @ 5) = 64 \quad \Rightarrow \quad 6 \# 4 \neq (14 @ 2) + (2 @ 5) \quad \text{(False)} \]

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#### Statement (iii): \( (6 @ 2) + (6 \# 3) = (12 \# 2) + (14 @ 12) - (2 @ 6) \)

1. **We already know \( 6 @ 2 = 16 \) and \( 12 \# 2 = 16 \)**.

2. **Calculate \( 6 \# 3 \)**:
- \( |6 + 3| = 9 \) (odd)
- Use the second case of \( \# \):
\[ 6 \# 3 = 2(6) @ 2(3) = 12 @ 6 \]
- Calculate \( 12 @ 6 \):
- \( |12 + 6| = 18 \) (even)
- Use the first case of \( @ \):
\[ 12 @ 6 = 3(12) - 6 = 36 - 6 = 30 \]

3. **Calculate \( 14 @ 12 \)**:
- \( |14 + 12| = 26 \) (even)
- Use the first case of \( @ \):
\[ 14 @ 12 = 3(14) - 12 = 42 - 12 = 30 \]

4. **Calculate \( 2 @ 6 \)**:
- \( |2 + 6| = 8 \) (even)
- Use the first case of \( @ \):
\[ 2 @ 6 = 3(2) - 6 = 6 - 6 = 0 \]

5. **Combine Results**:
\[ (6 @ 2) + (6 \# 3) = 16 + 30 = 46 \]
\[ (12 \# 2) + (14 @ 12) - (2 @ 6) = 16 + 30 - 0 = 46 \]

6. **Conclusion for Statement (iii)**:
\[ (6 @ 2) + (6 \# 3) = (12 \# 2) + (14 @ 12) - (2 @ 6) \quad \Rightarrow \quad \text{(True)} \]

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### Final Conclusion:
- Statement (i) is True.
- Statement (ii) is False.
- Statement (iii) is True.

The correct answer is **(i) and (iii) are true**.