We are tasked with finding the distance between the point $(8, 0, 0)$ and the line given by the parametric equations:

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

To find the distance between a point and a line in 3D space, we use the formula:

$d = \frac{| \mathbf{PQ} \cdot (\mathbf{v} \times \mathbf{PQ})|}{|\mathbf{v}|},$ where:

• $\mathbf{P} = (8, 0, 0)$ is the point,
• $\mathbf{Q}$ is a point on the line,
• $\mathbf{v}$ is the direction vector of the line.

Step 1: Direction vector of the line

From the parametric equations:

• The direction vector $\mathbf{v}$ of the line is $\langle 1, 2, 2 \rangle$, because these are the coefficients of $t$ in each parametric equation.

Step 2: Point on the line

For simplicity, we can choose $t = 0$, which gives us a point $\mathbf{Q}$ on the line: $\mathbf{Q} = (8, 0, 21\sqrt{5}).$

Step 3: Vector from $P$ to $Q$

The vector $\mathbf{PQ}$ is: $\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (8, 0, 21\sqrt{5}) - (8, 0, 0) = \langle 0, 0, 21\sqrt{5} \rangle.$

Step 4: Cross product $\mathbf{v} \times \mathbf{PQ}$

Now, compute the cross product: $\mathbf{v} \times \mathbf{PQ} = \langle 1, 2, 2 \rangle \times \langle 0, 0, 21\sqrt{5} \rangle.$

The determinant of the matrix will give us the cross product: