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.