The problem asks to find the distance from the point $(8, 0, 0)$ to the line given by the parametric equations:

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

Step 1: Identify the direction vector of the line.

From the parametric equations, the direction vector $\mathbf{v}$ of the line is the coefficients of $t$ in each equation:

$\mathbf{v} = (1, 2, 2).$

Step 2: Define the point on the line.

At $t = 0$, the point on the line is:

$P_0 = (8, 0, 6\sqrt{5}).$

Step 3: Use the formula for the distance from a point to a line.

The formula for the distance from a point $P_1$ to a line passing through a point $P_0$ with direction vector $\mathbf{v}$ is:

$d = \frac{\left| \mathbf{AP_0} \times \mathbf{v} \right|}{|\mathbf{v}|},$

where $\mathbf{AP_0}$ is the vector from the point $P_1$ to the point $P_0$.

• $P_1 = (8, 0, 0)$ (the point we are calculating the distance from),
• $P_0 = (8, 0, 6\sqrt{5})$,
• $\mathbf{v} = (1, 2, 2)$.

Step 4: Compute the vectors and the cross product.

Now, I'll calculate the distance using these parameters.The distance from the point $(8, 0, 0)$ to the given line is approximately $10$ units. This matches the correct answer from the choices provided.

Would you like further details on any step?

Here are 5 related questions for further practice:

1. How do you find the shortest distance between two skew lines?
2. What is the distance from a point to a plane?
3. How do you derive the equation of a plane given three points?
4. How is the dot product used in finding angles between vectors?
5. What is the geometric interpretation of the cross product?

Tip: In 3D geometry, the direction vector of a line plays a crucial role in determining distances and angles between lines and planes.