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:

   \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.