Let's solve the problem step by step.

Part (a): Finding the Equation of the Plane

To find the equation of the plane passing through three points $A(1, -1, 2)$, $B(1, 0, 1)$, and $C(-1, 0, 5)$, we need to follow these steps:

1. Find two vectors that lie on the plane: $\overrightarrow{AB} = B - A = (1 - 1, 0 - (-1), 1 - 2) = (0, 1, -1)$ $\overrightarrow{AC} = C - A = (-1 - 1, 0 - (-1), 5 - 2) = (-2, 1, 3)$

2. Find the normal vector to the plane by taking the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $\overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC}$

   \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -1 \\ -2 & 1 & 3 \end{vmatrix} = \mathbf{i}(1 \cdot 3 - (-1) \cdot 1) - \mathbf{j}(0 \cdot 3 - (-1) \cdot (-2)) + \mathbf{k}(0 \cdot 1 - 1 \cdot (-2))$$ $$\overrightarrow{n} = \mathbf{i}(3 + 1) - \mathbf{j}(0 - 2) + \mathbf{k}(0 + 2)$$ $$\overrightarrow{n} = 4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$$ So, the normal vector $$\overrightarrow{n}$$ is $$(4, -2, 2)$$.

3. Write the equation of the plane using the point-normal form of the plane equation: $4(x - 1) - 2(y + 1) + 2(z - 2) = 0$ Simplify the equation: $4x - 4 - 2y - 2 + 2z - 4 = 0$ $4x - 2y + 2z - 10 = 0$

Thus, the equation of the plane is: $4x - 2y + 2z = 10$

Part (b): Checking if Point $D(1, 2, 9)$ Belongs to the Plane and Finding the Distance if Not

1. Check if $D(1, 2, 9)$ lies on the plane: Substitute $D(1, 2, 9)$ into the plane equation $4x - 2y + 2z = 10$: $4(1) - 2(2) + 2(9) = 4 - 4 + 18 = 18 \neq 10$ Since the left side does not equal 10, point $D$ does not lie on the plane.

2. Find the distance from $D$ 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: $\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$ For the plane $4x - 2y + 2z - 10 = 0$, we have $A = 4$, $B = -2$, $C = 2$, and $D = -10$. The point $D$ is $(1, 2, 9)$. $\text{Distance} = \frac{|4(1) - 2(2) + 2(9) - 10|}{\sqrt{4^2 + (-2)^2 + 2^2}}$ $\text{Distance} = \frac{|4 - 4 + 18 - 10|}{\sqrt{16 + 4 + 4}}$ $\text{Distance} = \frac{|8|}{\sqrt{24}}$ $\text{Distance} = \frac{8}{2\sqrt{6}} = \frac{4}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$

Thus, the distance from point $D(1, 2, 9)$ to the plane is $\frac{2\sqrt{6}}{3}$.

Would you like any additional details or explanations? Here are some questions you might be interested in:

1. How do you compute a cross product in vector algebra?
2. What are the steps to find the normal vector to a plane?
3. Can you explain the point-normal form of a plane equation?
4. How do you determine if a point lies on a given plane?
5. What is the formula to calculate the distance from a point to a plane?

Tip: When working with planes and vectors, visualizing the geometric relationships can be extremely helpful in understanding the concepts.