To find the equation of a plane passing through three points, we need the following steps:

Step 1: Define the general equation of a plane

The equation of a plane in 3D space is of the form: $Ax + By + Cz + D = 0$ where $A$, $B$, and $C$ are the components of the normal vector to the plane, and $D$ is a constant.

Step 2: Find two vectors in the plane

Given three points $P_1(4,3,5)$, $P_2(1,2,0)$, and $P_3(1,0,2)$, we first compute two vectors that lie in the plane. These can be obtained by subtracting the coordinates of one point from the others:

1. Vector $\vec{v_1} = P_2 - P_1 = (1-4, 2-3, 0-5) = (-3, -1, -5)$
2. Vector $\vec{v_2} = P_3 - P_1 = (1-4, 0-3, 2-5) = (-3, -3, -3)$

Step 3: Compute the normal vector to the plane

The normal vector $\vec{n} = (A, B, C)$ to the plane is perpendicular to both $\vec{v_1}$ and $\vec{v_2}$. We find $\vec{n}$ by computing the cross product of $\vec{v_1}$ and $\vec{v_2}$:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -1 & -5 \\ -3 & -3 & -3 \end{vmatrix}$$ Expanding this determinant: $$\vec{n} = \mathbf{i} \left((-1)(-3) - (-5)(-3)\right) - \mathbf{j} \left((-3)(-3) - (-5)(-3)\right) + \mathbf{k} \left((-3)(-3) - (-3)(-1)\right)$$ $$\vec{n} = \mathbf{i} (3 - 15) - \mathbf{j} (9 - 15) + \mathbf{k} (9 - 3)$$ $$\vec{n} = \mathbf{i} (-12) - \mathbf{j} (-6) + \mathbf{k} (6)$$ $$\vec{n} = (-12, 6, 6)$$ Thus, the normal vector to the plane is $$\vec{n} = (-12, 6, 6)$$. ### Step 4: Write the equation of the plane Now that we have the normal vector $$(-12, 6, 6)$$, we can substitute these values into the general plane equation $$Ax + By + Cz + D = 0$$: $$-12(x - 4) + 6(y - 3) + 6(z - 5) = 0$$ Expanding this equation: $$-12x + 48 + 6y - 18 + 6z - 30 = 0$$ $$-12x + 6y + 6z = 0$$ Dividing the equation by 6 to simplify: $$-2x + y + z = 0$$ Thus, the equation of the plane is: $$\boxed{-2x + y + z = 0}$$ Let me know if you would like any further details or explanations! --- Here are 5 related questions to deepen your understanding: 1. How can you find the distance from a point to a plane using the plane equation? 2. What is the geometric interpretation of the cross product of two vectors? 3. How do you find the equation of a plane parallel to a given one? 4. How would the equation change if we translated the plane by a certain vector? 5. How do you check if a given point lies on the plane? **Tip:** The cross product of two vectors provides a perpendicular vector, which is useful for finding planes and lines in 3D geometry.