Let Q be the image of the point P(-2,1,-5) in the plane `3x-2y+2z+1=0`
Consider the following :
1. The coordinates of Q are (4,-3,-1).
2. PQ is of length more than 8 units.
3. The point (1,-1,-3) is the mid-point of the line segment PQ and lines on the given plane.
Which of the above statements are correct?

Options

A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3

Explanation

To solve the problem, we need to analyze the statements regarding the point \( Q \), which is the image of point \( P(-2, 1, -5) \) in the plane defined by the equation \( 3x - 2y + 2z + 1 = 0 \).

### Step 1: Find the Normal Vector of the Plane
The normal vector \( \vec{n} \) of the plane \( 3x - 2y + 2z + 1 = 0 \) can be derived from the coefficients of \( x, y, z \) in the equation. Thus,
\[ \vec{n} = (3, -2, 2). \]

### Step 2: Find the Foot of the Perpendicular from Point P to the Plane
To find the foot of the perpendicular from point \( P \) to the plane, we can use the formula for the foot of the perpendicular from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \):
\[ \left( x, y, z \right) = \left( x_0, y_0, z_0 \right) - \frac{A(x_0) + B(y_0) + C(z_0) + D}{A^2 + B^2 + C^2} \cdot (A, B, C). \]
Here, \( A = 3, B = -2, C = 2, D = 1 \), and \( (x_0, y_0, z_0) = (-2, 1, -5) \).

Calculating \( A(x_0) + B(y_0) + C(z_0) + D \):
\[ 3(-2) - 2(1) + 2(-5) + 1 = -6 - 2 - 10 + 1 = -17. \]

Calculating \( A^2 + B^2 + C^2 \):
\[ 3^2 + (-2)^2 + 2^2 = 9 + 4 + 4 = 17. \]

Now substituting into the formula:
\[ \left( x, y, z \right) = \left( -2, 1, -5 \right) - \frac{-17}{17} \cdot (3, -2, 2) = \left( -2, 1, -5 \right) + (3, -2, 2) = (1, -1, -3). \]

### Step 3: Find the Coordinates of Point Q
Since \( (1, -1, -3) \) is the foot of the perpendicular, the coordinates of point \( Q \) can be found by reflecting point \( P \) across this foot. The midpoint \( M \) of \( P \) and \( Q \) is given by:
\[ M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}, \frac{z_P + z_Q}{2} \right). \]
We know \( M = (1, -1, -3) \) and \( P = (-2, 1, -5) \). Let \( Q = (x_Q, y_Q, z_Q) \).

Setting up the equations:
1. \( \frac{-2 + x_Q}{2} = 1 \) leads to \( x_Q = 4 \).
2. \( \frac{1 + y_Q}{2} = -1 \) leads to \( y_Q = -3 \).
3. \( \frac{-5 + z_Q}{2} = -3 \) leads to \( z_Q = -1 \).

Thus, the coordinates of \( Q \) are \( (4, -3, -1) \).

### Step 4: Calculate the Length of PQ
Using the distance formula:
\[ PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2} = \sqrt{(4 - (-2))^2 + (-3 - 1)^2 + (-1 - (-5))^2}. \]
Calculating each term:
1. \( (4 + 2)^2 = 6^2 = 36 \).
2. \( (-3 - 1)^2 = (-4)^2 = 16 \).
3. \( (-1 + 5)^2 = 4^2 = 16 \).

Thus,
\[ PQ = \sqrt{36 + 16 + 16} = \sqrt{68} = 2\sqrt{17}. \]

### Step 5: Check the Length of PQ
To determine if \( PQ \) is more than 8 units, we check:
\[ 2\sqrt{17} > 8 \implies \sqrt{17} > 4 \implies 17 > 16, \]
which is true.

### Step 6: Verify if the Midpoint Lies on the Plane
We already know that the midpoint \( (1, -1, -3) \) satisfies the plane equation:
\[ 3(1) - 2(-1) + 2(-3) + 1 = 3 + 2 - 6 + 1 = 0. \]

### Conclusion
1. The coordinates of \( Q \) are \( (4, -3, -1) \) - **True**.
2. \( PQ \) is of length more than 8 units - **True**.
3. The point \( (1, -1, -3) \) is the midpoint of \( PQ \) and lies on the given plane - **True**.

Thus, all three statements are correct.