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The problem is asking to find the distance from the point \((8, 0, 0)\) to the line given by the parametric equations:

\[
x = 8 + t,\quad y = 2t,\quad z = 6\sqrt{5} + 2t
\]

To solve this, we can follow these steps:

1. **Direction Vector of the Line**: From the parametric equations, the direction vector of the line is obtained from the coefficients of \(t\):
   \[
   \mathbf{v} = \langle 1, 2, 2 \rangle
   \]

2. **Point on the Line**: One point on the line can be obtained by substituting \(t = 0\) into the parametric equations:
   \[
   P_0 = (8, 0, 6\sqrt{5})
   \]

3. **Vector from the Point to the Line**: The vector from the point \((8, 0, 0)\) to \(P_0 = (8, 0, 6\sqrt{5})\) is:
   \[
   \mathbf{u} = \langle 8 - 8, 0 - 0, 0 - 6\sqrt{5} \rangle = \langle 0, 0, -6\sqrt{5} \rangle
   \]

4. **Cross Product**: The distance from a point to a line in 3D can be found using the formula:
   \[
   d = \frac{\|\mathbf{u} \times \mathbf{v}\|}{\|\mathbf{v}\|}
   \]
   So we first need to compute the cross product \(\mathbf{u} \times \mathbf{v}\):
   \[
   \mathbf{u} \times \mathbf{v} = \langle 0, 0, -6\sqrt{5} \rangle \times \langle 1, 2, 2 \rangle
   \]
   Using the determinant formula for the cross product:
   \[
   \mathbf{u} \times \mathbf{v} = \langle (0 \cdot 2 - 0 \cdot 2), (0 \cdot 1 - (-6\sqrt{5}) \cdot 2), (0 \cdot 2 - 0 \cdot 1) \rangle = \langle 0, 12\sqrt{5}, 0 \rangle
   \]

5. **Magnitude of the Cross Product**: The magnitude of the cross product is:
   \[
   \|\mathbf{u} \times \mathbf{v}\| = \sqrt{0^2 + (12\sqrt{5})^2 + 0^2} = 12\sqrt{5}
   \]

6. **Magnitude of the Direction Vector**: The magnitude of the direction vector \(\mathbf{v}\) is:
   \[
   \|\mathbf{v}\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3
   \]

7. **Distance**: Finally, the distance is:
   \[
   d = \frac{12\sqrt{5}}{3} = 4\sqrt{5}
   \]

Thus, the distance from the point \((8, 0, 0)\) to the line is \(4\sqrt{5}\).

Would you like further explanation on any step? Here are some related questions to deepen your understanding:

1. How is the direction vector derived from parametric equations?
2. What is the geometrical significance of the cross product in this context?
3. How do you calculate the distance from a point to a line in 2D vs. 3D?
4. What are other methods to find the distance from a point to a line in 3D?
5. Can this method be extended to find the distance from a point to a plane?

**Tip**: When working with vectors, always remember that the cross product gives a vector perpendicular to the two original vectors, and its magnitude is related to the area of the parallelogram they form.

Find the distance from the point (8, 0, 0) to the line: x = 8 + t, y = 2t, z = 6√5 + 2t.

Learn how to calculate the distance from the point (8, 0, 0) to the line given by x = 8 + t, y = 2t, z = 6√5 + 2t using vector algebra. This detailed guide covers the use of direction vectors, cross products, and distance formulas in three-dimensional space. Suitable for college students or advanced high school studies in geometry and algebra.

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