Find the distance the point P(7, 1, 6), is to the plane through the three points
Q(5, 3, 4), R(7, -2, 3), and S(3, -2, 5).

To find the distance from the point \( P(7, 1, 6) \) to the plane determined by the points \( Q(5, 3, 4) \), \( R(7, -2, 3) \), and \( S(3, -2, 5) \), we can follow these steps:

### Step 1: Find two vectors that lie on the plane.
The vectors \( \overrightarrow{QR} \) and \( \overrightarrow{QS} \) represent two directions in the plane. They can be computed as:
\[
\overrightarrow{QR} = R - Q = (7 - 5, -2 - 3, 3 - 4) = (2, -5, -1)
\]
\[
\overrightarrow{QS} = S - Q = (3 - 5, -2 - 3, 5 - 4) = (-2, -5, 1)
\]

### Step 2: Compute the normal vector to the plane.
To find the normal vector to the plane, we compute the cross product of \( \overrightarrow{QR} \) and \( \overrightarrow{QS} \):
\[
\overrightarrow{n} = \overrightarrow{QR} \times \overrightarrow{QS}
\]
\[
\overrightarrow{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -5 & -1 \\ -2 & -5 & 1 \end{vmatrix}
\]
This gives us:
\[
\overrightarrow{n} = \mathbf{i} \left( (-5)(1) - (-5)(-1) \right) - \mathbf{j} \left( (2)(1) - (-1)(-2) \right) + \mathbf{k} \left( (2)(-5) - (-5)(-2) \right)
\]
\[
\overrightarrow{n} = \mathbf{i}( -5 - 5 ) - \mathbf{j}( 2 - 2 ) + \mathbf{k}( -10 - 10 )
\]
\[
\overrightarrow{n} = \mathbf{i}(-10) - \mathbf{k}(20)
\]
Thus, the normal vector is:
\[
\overrightarrow{n} = (-10, 0, -20)
\]

### Step 3: Equation of the plane.
The equation of the plane is given by:
\[
\overrightarrow{n} \cdot (x - x_1, y - y_1, z - z_1) = 0
\]
Using the point \( Q(5, 3, 4) \) and the normal vector \( (-10, 0, -20) \), we have:
\[
-10(x - 5) + 0(y - 3) - 20(z - 4) = 0
\]
Simplifying this equation:
\[
-10(x - 5) - 20(z - 4) = 0
\]
\[
-10x + 50 - 20z + 80 = 0
\]
\[
-10x - 20z + 130 = 0
\]
Thus, the equation of the plane is:
\[
10x + 20z = 130
\]

### Step 4: Distance from the point to the plane.
The distance from a point \( P(x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is given by the formula:
\[
\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
In this case, the equation of the plane is \( 10x + 20z - 130 = 0 \), so \( A = 10 \), \( B = 0 \), \( C = 20 \), and \( D = -130 \).

The coordinates of point \( P \) are \( (7, 1, 6) \), so we substitute into the formula:
\[
\text{Distance} = \frac{|10(7) + 0(1) + 20(6) - 130|}{\sqrt{10^2 + 0^2 + 20^2}}
\]
\[
\text{Distance} = \frac{|70 + 0 + 120 - 130|}{\sqrt{100 + 400}}
\]
\[
\text{Distance} = \frac{|60|}{\sqrt{500}}
\]
\[
\text{Distance} = \frac{60}{10\sqrt{5}}
\]
\[
\text{Distance} = \frac{6}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{6\sqrt{5}}{5}
\]

Thus, the distance from point \( P(7, 1, 6) \) to the plane is \( \frac{6\sqrt{5}}{5} \).

Would you like to go into more details or have any questions? Here are some related questions:

1. How can the cross product be used to find a normal vector to a plane?
2. What is the geometric significance of a plane's normal vector?
3. Why do we need two vectors to define a plane?
4. How is the formula for the distance between a point and a plane derived?
5. What are some alternative ways to describe the distance from a point to a plane?

**Tip:** Cross products are a valuable tool in vector geometry, particularly for finding perpendicular directions!

Find the distance the point P(7, 1, 6), is to the plane through the three points Q(5, 3, 4), R(7, -2, 3), and S(3, -2, 5).

Learn how to calculate the distance from the point P(7, 1, 6) to a plane determined by three points Q(5, 3, 4), R(7, -2, 3), and S(3, -2, 5). This solution covers key concepts in vector geometry, including finding the normal vector using cross product and applying the distance formula. Suitable for high school students in Grades 10-12.

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