Find the distance the point P(8,-5,0) is to the line through the two points
Q(4,-3,2), and R(6,-6,3).

To find the distance from the point \( P(8, -5, 0) \) to the line through points \( Q(4, -3, 2) \) and \( R(6, -6, 3) \), we need to use the formula for the distance between a point and a line in 3D space.

### Step-by-Step Solution:

1. **Find the direction vector of the line:**

   The direction vector \( \mathbf{v} \) of the line through points \( Q \) and \( R \) is given by:
   \[
   \mathbf{v} = \overrightarrow{QR} = (R_x - Q_x, R_y - Q_y, R_z - Q_z)
   \]

   Substituting the coordinates of \( Q(4, -3, 2) \) and \( R(6, -6, 3) \):

   \[
   \mathbf{v} = (6 - 4, -6 + 3, 3 - 2) = (2, -3, 1)
   \]

2. **Find the vector from point \( Q \) to point \( P \):**

   The vector \( \mathbf{u} \) from point \( Q \) to point \( P \) is:

   \[
   \mathbf{u} = \overrightarrow{QP} = (P_x - Q_x, P_y - Q_y, P_z - Q_z)
   \]

   Substituting the coordinates of \( P(8, -5, 0) \) and \( Q(4, -3, 2) \):

   \[
   \mathbf{u} = (8 - 4, -5 + 3, 0 - 2) = (4, -2, -2)
   \]

3. **Compute the cross product of \( \mathbf{u} \) and \( \mathbf{v} \):**

   The cross product \( \mathbf{u} \times \mathbf{v} \) is:

   \[
   \mathbf{u} \times \mathbf{v} = 
   \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   4 & -2 & -2 \\
   2 & -3 & 1
   \end{vmatrix}
   \]

   Calculating the determinant:

   \[
   \mathbf{u} \times \mathbf{v} = \mathbf{i} \left( (-2)(1) - (-2)(-3) \right) - \mathbf{j} \left( (4)(1) - (-2)(2) \right) + \mathbf{k} \left( (4)(-3) - (-2)(2) \right)
   \]

   Simplifying:

   \[
   \mathbf{u} \times \mathbf{v} = \mathbf{i} \left( -2 - 6 \right) - \mathbf{j} \left( 4 + 4 \right) + \mathbf{k} \left( -12 + 4 \right)
   \]

   \[
   \mathbf{u} \times \mathbf{v} = \mathbf{i} (-8) - \mathbf{j} (8) + \mathbf{k} (-8)
   \]

   \[
   \mathbf{u} \times \mathbf{v} = (-8, -8, -8)
   \]

4. **Find the magnitude of the cross product:**

   The magnitude of \( \mathbf{u} \times \mathbf{v} \) is:

   \[
   \left| \mathbf{u} \times \mathbf{v} \right| = \sqrt{(-8)^2 + (-8)^2 + (-8)^2}
   \]

   \[
   \left| \mathbf{u} \times \mathbf{v} \right| = \sqrt{64 + 64 + 64} = \sqrt{192} = 8\sqrt{3}
   \]

5. **Find the magnitude of the direction vector \( \mathbf{v} \):**

   The magnitude of \( \mathbf{v} = (2, -3, 1) \) is:

   \[
   \left| \mathbf{v} \right| = \sqrt{2^2 + (-3)^2 + 1^2}
   \]

   \[
   \left| \mathbf{v} \right| = \sqrt{4 + 9 + 1} = \sqrt{14}
   \]

6. **Compute the distance from point \( P \) to the line:**

   The distance \( d \) from point \( P \) to the line is given by:

   \[
   d = \frac{\left| \mathbf{u} \times \mathbf{v} \right|}{\left| \mathbf{v} \right|}
   \]

   Substituting the values:

   \[
   d = \frac{8\sqrt{3}}{\sqrt{14}}
   \]

   Simplify by rationalizing the denominator:

   \[
   d = \frac{8\sqrt{3} \cdot \sqrt{14}}{14} = \frac{8\sqrt{42}}{14} = \frac{4\sqrt{42}}{7}
   \]

### Final Answer:

The distance from point \( P(8, -5, 0) \) to the line through points \( Q(4, -3, 2) \) and \( R(6, -6, 3) \) is:

\[
\boxed{\frac{4\sqrt{42}}{7}}
\]

Would you like more details on any step, or do you have any questions?

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### Related Questions:

1. How do you compute the cross product of two vectors?
2. What is the formula for finding the distance from a point to a line in 3D space?
3. How do you find the direction vector of a line given two points?
4. What are some practical applications of finding the distance from a point to a line?
5. How can the cross product be used in physics and engineering?

**Tip:** When solving distance problems, understanding vector operations like dot and cross products is essential.

Find the distance the point P(8,-5,0) is to the line through the two points Q(4,-3,2), and R(6,-6,3).

Discover how to calculate the distance from point P(8, -5, 0) to the line passing through points Q(4, -3, 2) and R(6, -6, 3) using vector geometry and cross product. This step-by-step guide is suitable for advanced high school or college students.

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