To find the implicit equation of a plane passing through three points, we use the following general equation of a plane:

$Ax + By + Cz + D = 0$

Step-by-step process:

1. Label the given points: $P_1 = (5, 2, -2), \quad P_2 = (0, 7, -5), \quad P_3 = (4, 1, -3)$

2. Find two vectors that lie on the plane by subtracting coordinates of the points: $\vec{v_1} = P_2 - P_1 = (0-5, 7-2, -5 - (-2)) = (-5, 5, -3)$ $\vec{v_2} = P_3 - P_1 = (4-5, 1-2, -3 - (-2)) = (-1, -1, -1)$

3. Find the normal vector to the plane by taking the cross product of $\vec{v_1}$ and $\vec{v_2}$: $\vec{n} = \vec{v_1} \times \vec{v_2}$ The cross product $\vec{v_1} \times \vec{v_2}$ is computed as: $\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 5 & -3 \\ -1 & -1 & -1 \end{vmatrix}$ Expanding this determinant: $\vec{n} = \mathbf{i} \left( 5(-1) - (-3)(-1) \right) - \mathbf{j} \left( -5(-1) - (-3)(-1) \right) + \mathbf{k} \left( -5(-1) - 5(-1) \right)$ $\vec{n} = \mathbf{i}(-5 - 3) - \mathbf{j}(5 - 3) + \mathbf{k}(5 - 5)$ $\vec{n} = \mathbf{i}(-8) - \mathbf{j}(2) + \mathbf{k}(0)$ $\vec{n} = (-8, -2, 0)$ So, the normal vector $\vec{n} = (-8, -2, 0)$.

4. Write the equation of the plane:

   Using the general form of the plane equation $Ax + By + Cz + D = 0$ and the normal vector $\vec{n} = (-8, -2, 0)$, the equation becomes: $-8(x - 5) - 2(y - 2) + 0(z + 2) = 0$ Simplifying: $-8x + 40 - 2y + 4 = 0$ $-8x - 2y + 44 = 0$ So, the implicit equation of the plane is: $8x + 2y = 44$

Let me know if you'd like more details or clarifications.

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Here are 5 related questions:

1. How do you find the cross product of two vectors in general?
2. What is the geometric meaning of the normal vector to a plane?
3. Can you derive the parametric equation of the plane using the same points?
4. What are the conditions for three points to lie on the same plane?
5. How do you find the distance from a point to a plane using the plane's equation?

Tip: When calculating cross products, using determinants can help simplify the process.