The image you've uploaded appears to contain questions related to lines, planes, and distances in 3D geometry. Let's go through the questions one by one:

1. Intersection of a line with the $xy$-plane: The line passes through the points $A(1, -2, 4)$ and $B(2, 3, 2)$. To find the point where this line intersects the $xy$-plane, set the $z$-coordinate equal to 0 (since the $xy$-plane has $z = 0$).

   The parametric equations for the line can be written as: $x = 1 + t(2 - 1), \quad y = -2 + t(3 - (-2)), \quad z = 4 + t(2 - 4)$ Substitute $z = 0$ and solve for $t$ to find the coordinates where the line intersects the plane.

2. Equation of the plane with normal vector: The equation of a plane is given by: $n \cdot (x - x_0, y - y_0, z - z_0) = 0$ where $n = (5, -3, 3)$ is the normal vector, and the point $P(2, 2, -2)$ lies on the plane. Plugging in the values, we can find the equation of the plane.

3. Plane passing through three points: The points are $(1, 1, 1), (2, 0, 3), (-1, 4, 2)$. To find the equation of the plane, you need to:

   • Find two vectors on the plane by subtracting coordinates of two points.
   • Compute the cross product of these two vectors to find the normal vector.
   • Use the normal vector and one of the points to write the equation of the plane.

4. Distance from a point to a plane: The point is $(-5, -3, 5)$ and the plane equation is $3x - 4z + 17 = 0$. To find the distance, use the formula: $\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$ where $(A, B, C)$ is the normal vector of the plane, and $(x_1, y_1, z_1)$ is the point.

Would you like me to solve one of these in more detail or help with any specific part of the questions?