Let P denotes the plane consisting of all points that are equidistant from the points `A(-4,2,1)` and `B(2,-4,3)` and Q be the plane, `x-y+cz=1` where `c in R`.
If the angle between the planes P and Q is `45^(@)` then the product of all possible values of c is

Options

A. `-17`
B. `-2`
C. 17
D. `24//27`

Explanation

To solve the problem, we need to find the equation of the plane \( P \) that consists of all points equidistant from points \( A(-4, 2, 1) \) and \( B(2, -4, 3) \). Then, we will find the angle between this plane \( P \) and the plane \( Q: x - y + cz = 1 \) and determine the product of all possible values of \( c \) when the angle is \( 45^\circ \).

### Step 1: Find the equation of the plane \( P \)

The plane \( P \) is defined as the locus of points equidistant from points \( A \) and \( B \). The equation of the plane can be derived from the condition that the distance from any point \( (x, y, z) \) to \( A \) is equal to the distance from that point to \( B \).

The distance from \( (x, y, z) \) to \( A \) is given by:
\[ \sqrt{(x + 4)^2 + (y - 2)^2 + (z - 1)^2} \]

The distance from \( (x, y, z) \) to \( B \) is given by:
\[ \sqrt{(x - 2)^2 + (y + 4)^2 + (z - 3)^2} \]

Setting these distances equal gives us:
\[ (x + 4)^2 + (y - 2)^2 + (z - 1)^2 = (x - 2)^2 + (y + 4)^2 + (z - 3)^2 \]

### Step 2: Expand and simplify the equation

Expanding both sides:
\[ (x^2 + 8x + 16) + (y^2 - 4y + 4) + (z^2 - 2z + 1) = (x^2 - 4x + 4) + (y^2 + 8y + 16) + (z^2 - 6z + 9) \]

Combining like terms:
\[ 8x + 16 - 4y + 4 - 2z + 1 = -4x + 4 + 8y + 16 - 6z + 9 \]

This simplifies to:
\[ 12x - 12y + 4z + 6 = 0 \]

Dividing through by 6 gives:
\[ 2x - 2y + \frac{2}{3}z + 1 = 0 \]

### Step 3: Find the normal vector of plane \( P \)

The normal vector \( \mathbf{n_1} \) of plane \( P \) is given by the coefficients of \( x, y, z \):
\[ \mathbf{n_1} = (2, -2, \frac{2}{3}) \]

### Step 4: Find the normal vector of plane \( Q \)

The normal vector \( \mathbf{n_2} \) of plane \( Q: x - y + cz = 1 \) is:
\[ \mathbf{n_2} = (1, -1, c) \]

### Step 5: Calculate the angle between the planes

The angle \( \theta \) between the two planes is given by:
\[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} \]

For \( \theta = 45^\circ \), we have:
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \]

Calculating the dot product:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 2 \cdot 1 + (-2) \cdot (-1) + \frac{2}{3} \cdot c = 2 + 2 + \frac{2}{3}c = 4 + \frac{2}{3}c \]

Calculating the magnitudes:
\[ |\mathbf{n_1}| = \sqrt{2^2 + (-2)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{4 + 4 + \frac{4}{9}} = \sqrt{\frac{36 + 36 + 4}{9}} = \sqrt{\frac{76}{9}} = \frac{\sqrt{76}}{3} \]

\[ |\mathbf{n_2}| = \sqrt{1^2 + (-1)^2 + c^2} = \sqrt{1 + 1 + c^2} = \sqrt{2 + c^2} \]

### Step 6: Set up the equation

Substituting into the angle formula:
\[ \frac{4 + \frac{2}{3}c}{\frac{\sqrt{76}}{3} \cdot \sqrt{2 + c^2}} = \frac{1}{\sqrt{2}} \]

Cross-multiplying gives:
\[ (4 + \frac{2}{3}c) \sqrt{2} = \frac{\sqrt{76}}{3} \cdot \sqrt{2 + c^2} \]

### Step 7: Square both sides to eliminate the square root

Squaring both sides leads to a quadratic equation in \( c \). After simplification, we will find the values of \( c \).

### Step 8: Solve the quadratic equation

The resulting quadratic equation can be solved using the quadratic formula:
\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 9: Find the product of the roots

Using Vieta's formulas, the product of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \frac{c}{a} \).