To find the equation of a plane, we need two things:

1. A point on the plane.
2. A normal vector to the plane.

1. Point on the Plane

The plane passes through the point $(1, 0, 2)$.

2. Normal Vector

The plane is perpendicular to the line, so the direction vector of the line will be the normal vector to the plane.

The parametric equations of the line are:

$x = 4t, \quad y = 1 - t, \quad z = 6 + 6t$

The direction vector of the line is given by the coefficients of $t$:

$\vec{v} = \langle 4, -1, 6 \rangle$

Since the plane is perpendicular to this line, the vector $\vec{v} = \langle 4, -1, 6 \rangle$ will be the normal vector to the plane.

3. Equation of the Plane

The general form of the equation of a plane is:

$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

where $(x_0, y_0, z_0)$ is a point on the plane, and $\langle a, b, c \rangle$ is the normal vector.

Using the point $(1, 0, 2)$ and the normal vector $\langle 4, -1, 6 \rangle$, we plug these into the equation of the plane:

$4(x - 1) - 1(y - 0) + 6(z - 2) = 0$

4. Simplify the Equation

Now, simplify the expression:

$4(x - 1) - y + 6(z - 2) = 0$

Expanding:

$4x - 4 - y + 6z - 12 = 0$

$4x - y + 6z - 16 = 0$

Final Equation of the Plane:

The equation of the plane is:

$4x - y + 6z = 16$

Would you like further details or clarifications on any step?

Here are 5 related questions to explore further:

1. How would the equation change if the normal vector were different?
2. What if the point through which the plane passes was different?
3. How do you find the distance from a point to this plane?
4. How can you find the line of intersection of two planes?
5. What if the line was parallel to the plane instead of perpendicular?

Tip: When finding the equation of a plane, make sure you have both the normal vector and a point on the plane for accurate results.