Consider the plane `pi:x+y-2z=3` and two points `P(2,1,6)` and `Q(6,5,-2)`.
Statement 1: PQ is parallel to the normal to the plane.
Statement 2: `Q` is the image of point `P` in he plane `pi`

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze the two statements regarding the plane \( \pi: x + y - 2z = 3 \) and the points \( P(2, 1, 6) \) and \( Q(6, 5, -2) \).

### Step 1: Determine the vector \( \overrightarrow{PQ} \)

The vector \( \overrightarrow{PQ} \) can be calculated as follows:
\[ \overrightarrow{PQ} = Q - P = (6 - 2, 5 - 1, -2 - 6) = (4, 4, -8) \]

### Step 2: Identify the normal vector of the plane

The normal vector \( \vec{n} \) of the plane \( \pi: x + y - 2z = 3 \) is given by the coefficients of \( x, y, z \) in the equation of the plane:
\[ \vec{n} = (1, 1, -2) \]

### Step 3: Check if \( \overrightarrow{PQ} \) is parallel to the normal vector

To check if \( \overrightarrow{PQ} \) is parallel to \( \vec{n} \), we can express \( \overrightarrow{PQ} \) as a scalar multiple of \( \vec{n} \):
\[ \overrightarrow{PQ} = (4, 4, -8) = 4(1, 1, -2) = 4\vec{n} \]
Since \( \overrightarrow{PQ} \) is a scalar multiple of the normal vector, we conclude that \( \overrightarrow{PQ} \) is parallel to \( \vec{n} \).

**Conclusion for Statement 1:** True

### Step 4: Find the midpoint \( R \) of segment \( PQ \)

The midpoint \( R \) of points \( P \) and \( Q \) is calculated as follows:
\[ R = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = \left( \frac{2 + 6}{2}, \frac{1 + 5}{2}, \frac{6 - 2}{2} \right) = \left( 4, 3, 2 \right) \]

### Step 5: Check if the midpoint \( R \) lies on the plane \( \pi \)

To verify if point \( R(4, 3, 2) \) lies on the plane, substitute \( x = 4 \), \( y = 3 \), and \( z = 2 \) into the plane equation:
\[ 4 + 3 - 2(2) = 4 + 3 - 4 = 3 \]
Since the left-hand side equals the right-hand side of the equation \( 3 = 3 \), point \( R \) lies on the plane.

**Conclusion for Statement 2:** True, and since \( R \) is the midpoint of \( P \) and \( Q \), \( Q \) is indeed the image of \( P \) in the plane.

### Final Conclusion

Both statements are true:
- Statement 1: \( PQ \) is parallel to the normal to the plane. (True)
- Statement 2: \( Q \) is the image of point \( P \) in the plane \( \pi \). (True)

However, Statement 2 does not serve as an explanation for Statement 1.