if `p^(2)+2q^(2)+5r^(2)+2pq-4qr+6p+6q-2r+10le0` for `p,q,r in R` then identify correct option?

Options

A. `p+q^(2)+r^(2)=0`
B. `2q+r+p=0`
C. `r^(p)=3q+p`
D. `2^(3q+p)=p^(2)+r-2q`

Explanation

To solve the inequality \( p^2 + 2q^2 + 5r^2 + 2pq - 4qr + 6p + 6q - 2r + 10 \leq 0 \) for \( p, q, r \in \mathbb{R} \), we will rewrite the expression and analyze it step by step.

### Step 1: Rewrite the expression
We start with the inequality:
\[ p^2 + 2q^2 + 5r^2 + 2pq - 4qr + 6p + 6q - 2r + 10 \leq 0 \]
We can rearrange and group the terms to make it easier to analyze.

### Step 2: Complete the square
We can complete the square for the quadratic terms. Let's break down the terms:
- For \( p^2 + 2pq + 2q^2 \), we can complete the square:
\[ p^2 + 2pq + q^2 + q^2 = (p + q)^2 + q^2 \]
- For \( 5r^2 - 4qr - 2r \), we can rewrite it as:
\[ 5r^2 - 4qr - 2r = 5r^2 - (4q + 2)r \]
To complete the square, we can factor out 5:
\[ 5\left(r^2 - \frac{(4q + 2)}{5}r\right) \]
Now, we complete the square inside the parentheses.

### Step 3: Analyze the completed squares
After completing the squares, we rewrite the entire expression:
\[ (p + q + 3)^2 + (q - 2r)^2 + (r - 1)^2 \leq 0 \]
Since squares of real numbers are always non-negative, the only way for this sum to be less than or equal to zero is if each square is equal to zero.

### Step 4: Set each square to zero
1. \( (p + q + 3)^2 = 0 \) implies \( p + q + 3 = 0 \) or \( p + q = -3 \).
2. \( (q - 2r)^2 = 0 \) implies \( q - 2r = 0 \) or \( q = 2r \).
3. \( (r - 1)^2 = 0 \) implies \( r - 1 = 0 \) or \( r = 1 \).

### Step 5: Solve for \( p \), \( q \), and \( r \)
From \( r = 1 \):
- Substitute \( r = 1 \) into \( q = 2r \):
\[ q = 2 \cdot 1 = 2 \]
- Substitute \( q = 2 \) into \( p + q = -3 \):
\[ p + 2 = -3 \implies p = -5 \]

Thus, we have:
\[ p = -5, \quad q = 2, \quad r = 1 \]

### Step 6: Check the options
Now we will check which of the provided options are satisfied by \( p = -5 \), \( q = 2 \), and \( r = 1 \).

1. **Option 1**: \( p + q^2 + r^2 = 0 \)
\[ -5 + 2^2 + 1^2 = -5 + 4 + 1 = 0 \quad \text{(True)} \]

2. **Option 2**: \( 2q + r + p = 0 \)
\[ 2 \cdot 2 + 1 - 5 = 4 + 1 - 5 = 0 \quad \text{(True)} \]

3. **Option 3**: \( r^p = 3q + p \)
\[ 1^{-5} = 3 \cdot 2 - 5 \quad 1 = 6 - 5 = 1 \quad \text{(True)} \]

4. **Option 4**: \( 2^{3q + p} = p^2 + r - 2q \)
\[ 2^{3 \cdot 2 - 5} = (-5)^2 + 1 - 2 \cdot 2 \quad 2^{6 - 5} = 25 + 1 - 4 \quad 2^1 = 22 \quad \text{(False)} \]

### Final Answer
The correct options are:
- Option 1: \( p + q^2 + r^2 = 0 \)
- Option 2: \( 2q + r + p = 0 \)
- Option 3: \( r^p = 3q + p \)