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We need to calculate the distance from the point \( (8, 0, 0) \) to the line given by the parametric equations:

\[
x = 8 + t, \quad y = 2t, \quad z = 21\sqrt{5} + 2t.
\]

### Step-by-step solution:

1. **Direction vector of the line:**

   The direction vector of the line can be extracted from the parametric equations. It is given by the coefficients of \( t \):

   \[
   \vec{d} = \langle 1, 2, 2 \rangle.
   \]

2. **Point on the line:**

   When \( t = 0 \), the point on the line is:

   \[
   P_1 = (8, 0, 21\sqrt{5}).
   \]

3. **Vector from the point \( (8, 0, 0) \) to the point on the line \( P_1 \):**

   Let \( P_2 = (8, 0, 0) \) be the given point. The vector from \( P_2 \) to \( P_1 \) is:

   \[
   \vec{v} = P_1 - P_2 = \langle 8 - 8, 0 - 0, 21\sqrt{5} - 0 \rangle = \langle 0, 0, 21\sqrt{5} \rangle.
   \]

4. **Cross product of \( \vec{v} \) and \( \vec{d} \):**

   We need the cross product \( \vec{v} \times \vec{d} \) to find the perpendicular distance. The cross product of \( \langle 0, 0, 21\sqrt{5} \rangle \) and \( \langle 1, 2, 2 \rangle \) is:

   \[
   \vec{v} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 21\sqrt{5} \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(0 \cdot 2 - 21\sqrt{5} \cdot 2) - \hat{j}(0 \cdot 2 - 21\sqrt{5} \cdot 1) + \hat{k}(0 \cdot 2 - 0 \cdot 1)
   \]

   Simplifying, we get:

   \[
   \vec{v} \times \vec{d} = \langle -42\sqrt{5}, -21\sqrt{5}, 0 \rangle.
   \]

5. **Magnitude of the cross product:**

   The magnitude of \( \vec{v} \times \vec{d} \) is:

   \[
   |\vec{v} \times \vec{d}| = \sqrt{(-42\sqrt{5})^2 + (-21\sqrt{5})^2 + 0^2} = \sqrt{1764 \cdot 5 + 441 \cdot 5} = \sqrt{8820 + 2205} = \sqrt{11025} = 105.
   \]

6. **Magnitude of the direction vector:**

   The magnitude of \( \vec{d} = \langle 1, 2, 2 \rangle \) is:

   \[
   |\vec{d}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.
   \]

7. **Distance from the point to the line:**

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

   \[
   D = \frac{|\vec{v} \times \vec{d}|}{|\vec{d}|} = \frac{105}{3} = 35.
   \]

Thus, the distance from the point \( (8, 0, 0) \) to the line is **35**.

Let me know if you need any further clarifications!

### Related Questions:

1. How do you find the distance from a point to a plane?
2. What is the geometrical interpretation of the cross product?
3. How can you find the shortest distance between two skew lines?
4. What role does the direction vector play in parametric equations?
5. Can you derive the distance formula for two points in 3D space?

### Tip:

Always check that you're working with the correct direction vector for a line when finding distances or projections!

Find the distance from the point (8, 0, 0) to the line x = 8 + t, y = 2t, z = 21√5 + 2t.

Learn how to calculate the distance from the point (8, 0, 0) to a line defined by parametric equations. This solution uses vector geometry, the cross product, and distance formulas, and is suitable for Grade 11-12 students.

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