This problem involves analyzing a pyramid with given vertices, solving for the equation of a plane defined by three of its vertices, and determining the height of the pyramid from the fourth vertex to that plane. Here's how we can approach the solution:

Given:

• Vertices of the pyramid:
  • $A(3, 3, 5)$
  • $B(1, 1, -1)$
  • $C(-3, 3, 2)$
  • $D(1, 1, 0)$

We need to:

1. Find the equation of the plane containing points $A$, $B$, and $C$.
2. Compute the perpendicular distance from point $D$ to this plane, which will be the height $h_D$ of the pyramid.

Step 1: Find the equation of the plane $\text{ABC}$

The general equation of a plane is: $ax + by + cz + d = 0$ We need to find $a$, $b$, $c$, and $d$.

1. Determine the normal vector to the plane: Use two vectors lying on the plane: $\vec{AB} = (B_x - A_x, B_y - A_y, B_z - A_z) = (1 - 3, 1 - 3, -1 - 5) = (-2, -2, -6)$ $\vec{AC} = (C_x - A_x, C_y - A_y, C_z - A_z) = (-3 - 3, 3 - 3, 2 - 5) = (-6, 0, -3)$ The cross product $\vec{AB} \times \vec{AC}$ gives the normal vector: \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -2 & -6 \\ -6 & 0 & -3 \end{vmatrix}$$ Expand the determinant: $$\vec{n} = \mathbf{i}((-2)(-3) - (-6)(0)) - \mathbf{j}((-2)(-3) - (-6)(-6)) + \mathbf{k}((-2)(0) - (-6)(-2))$$ $$\vec{n} = \mathbf{i}(6) - \mathbf{j}(18 - 36) + \mathbf{k}(-12)$$ $$\vec{n} = (6, -18, -12)$$ Simplify: $$\vec{n} = (1, -3, -2) \quad \text{(dividing by the GCD 6)}.$$
2. Plane equation: Substitute $(1, -3, -2)$ as $(a, b, c)$, and use point $A(3, 3, 5)$ to solve for $d$: $1(x) - 3(y) - 2(z) + d = 0$ Substitute $A(3, 3, 5)$: $1(3) - 3(3) - 2(5) + d = 0$ $3 - 9 - 10 + d = 0$ $d = 16$ The equation of the plane is: $x - 3y - 2z + 16 = 0$

Step 2: Distance from $D(1, 1, 0)$ to the plane

The distance from a point $(x_1, y_1, z_1)$ to a plane $ax + by + cz + d = 0$ is given by: $\text{Distance} = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$ Substitute $D(1, 1, 0)$, $a = 1$, $b = -3$, $c = -2$, $d = 16$: $\text{Distance} = \frac{|1(1) - 3(1) - 2(0) + 16|}{\sqrt{1^2 + (-3)^2 + (-2)^2}}$ $\text{Distance} = \frac{|1 - 3 + 0 + 16|}{\sqrt{1 + 9 + 4}}$ $\text{Distance} = \frac{|14|}{\sqrt{14}}$ $\text{Distance} = \sqrt{14}$

Thus, the height $h_D$ of the pyramid is $\sqrt{14}$.

Final Answer:

1. Equation of the plane $\text{ABC}$: $x - 3y - 2z + 16 = 0$
2. Height of the pyramid $h_D$: $\sqrt{14}$

Would you like further clarification or a step-by-step breakdown of any part?

Here are five related questions:

1. How do you calculate the volume of the pyramid using this height?
2. How can we determine if a point lies on the plane?
3. What are alternative methods to find the normal vector of a plane?
4. How can the cross-product magnitude help determine the area of a triangle on a plane?
5. How would you rewrite the plane equation if the GCD were not used?

Tip: Always verify your cross-product calculations to avoid errors in determining the normal vector.