It looks like you need help with three problems involving planes and lines in 3D space.

1. Equation of a plane: You are given a point (2, 3, 6) and a normal vector (6, 9, 9), and you need to find the equation of the plane.

2. Equation of a plane: You need to find the equation of a plane passing through the point (10, -1, 5) and containing the line described by the symmetric equations.

3. Intersection of a line with a plane: You need to find the point where the line defined by the parametric equations $x = 3 + 2t, y = 3t, z = 5 - 3t$ intersects the plane defined by the equation $x + 3y - z + 2 = 0$.

Let's solve them one by one.

1. Equation of the Plane

Given:

• Point: $(2, 3, 6)$
• Normal vector: $\mathbf{n} = 6i + 9j + 9k$

The general form of the equation of a plane is: $n_1(x - x_1) + n_2(y - y_1) + n_3(z - z_1) = 0$ Where $(x_1, y_1, z_1)$ is a point on the plane, and $(n_1, n_2, n_3)$ is the normal vector.

Substitute $(x_1, y_1, z_1) = (2, 3, 6)$ and $\mathbf{n} = (6, 9, 9)$: $6(x - 2) + 9(y - 3) + 9(z - 6) = 0$ Simplify: $6x - 12 + 9y - 27 + 9z - 54 = 0$ $6x + 9y + 9z = 93$ This is the equation of the plane.

2. Equation of the Plane Containing a Line

You are given the point $(10, -1, 5)$ and a line with symmetric equations: $\frac{x - 10}{6} = \frac{y + 1}{1} = \frac{z - 5}{2}$

From these symmetric equations, you can obtain the direction vector of the line as $\mathbf{d} = (6, 1, 2)$.

You will need another direction vector, which is not provided directly, to fully define the plane. You can form this vector by using a point on the line and the given point $(10, -1, 5)$, or other available information.

3. Intersection of a Line and a Plane

Substitute the parametric equations of the line into the equation of the plane. The parametric equations are: $x = 3 + 2t, \quad y = 3t, \quad z = 5 - 3t$ Substitute into the plane equation $x + 3y - z + 2 = 0$: $(3 + 2t) + 3(3t) - (5 - 3t) + 2 = 0$ Simplify: $3 + 2t + 9t - 5 + 3t + 2 = 0$ $(2t + 9t + 3t) + (3 - 5 + 2) = 0$ $14t = 0$ So, $t = 0$.

Substitute $t = 0$ into the parametric equations to find the coordinates: $x = 3 + 2(0) = 3, \quad y = 3(0) = 0, \quad z = 5 - 3(0) = 5$ Thus, the point of intersection is $(3, 0, 5)$.

Would you like any further details on these problems or have additional questions?

Here are 5 related questions to expand on the topic:

1. How do you find the equation of a plane when the normal vector is not provided?
2. What is the geometric interpretation of the normal vector in the context of a plane?
3. How do you calculate the angle between two planes using their normal vectors?
4. How can you determine the distance from a point to a plane?
5. How can you determine if two planes are parallel or perpendicular?

Tip: To check if two planes are parallel, see if their normal vectors are scalar multiples of each other.