For every `a, b in N`
`a" @ "b=a^(2)" when "(a+b)" is even"`
`=a^(2)-b^(2)" when "(a+b)" is odd"`
`a#b=b^(2)" when "axxb" is odd"`
`=a^(2)-b^(2)" when "axxb" is even"`
Find the value of `(p#q)-(q" @ "p)-(q" @ "p),p` is even and q is odd :

Options

A. 0
B. `2(p^(2)+q^(2))`
C. `3(p^(2)-q^(2))`
D. none of these

Answer

To solve the problem, we need to evaluate the expression \((p \# q) - (q \, @ \, p) - (q \, @ \, p)\), given that \(p\) is even and \(q\) is odd.

### Step 1: Calculate \(p \# q\)

According to the definition provided:
- \(a \# b = b^2\) when \(a \times b\) is odd.
- \(a \# b = a^2 - b^2\) when \(a \times b\) is even.

Since \(p\) is even and \(q\) is odd, the product \(p \times q\) is even (even multiplied by odd is even). Therefore, we use the second case:

\[ p \# q = p^2 - q^2 \]

### Step 2: Calculate \(q \, @ \, p\)

According to the definition provided:
- \(a \, @ \, b = a^2\) when \(a + b\) is even.
- \(a \, @ \, b = a^2 - b^2\) when \(a + b\) is odd.

Since \(p\) is even and \(q\) is odd, the sum \(p + q\) is odd (even plus odd is odd). Therefore, we use the second case:

\[ q \, @ \, p = q^2 - p^2 \]

### Step 3: Substitute into the expression

Now we substitute the values we found into the expression:

\[ (p \# q) - (q \, @ \, p) - (q \, @ \, p) = (p^2 - q^2) - (q^2 - p^2) - (q^2 - p^2) \]

### Step 4: Simplify the expression

First, simplify \((q^2 - p^2)\):

\[ = (p^2 - q^2) - (q^2 - p^2) - (q^2 - p^2) \]

This simplifies to:

\[ = (p^2 - q^2) - (q^2 - p^2) - (q^2 - p^2) = (p^2 - q^2) - (q^2 - p^2) - (q^2 - p^2) = (p^2 - q^2) - 2(q^2 - p^2) \]

Now, combine like terms:

\[ = p^2 - q^2 - 2q^2 + 2p^2 = 3p^2 - 3q^2 = 3(p^2 - q^2) \]

### Final Answer

Thus, the final value is:

\[ 3(p^2 - q^2) \]