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To find the equation of a plane passing through three points, \( P(-4, -1, -2) \), \( Q(5, 2, 8) \), and \( R(-1, 0, 5) \), we can proceed with the following steps:

1. **Calculate Two Direction Vectors:**
   - Vector \( \vec{PQ} \) from \( P \) to \( Q \):
     \[
     \vec{PQ} = Q - P = (5 - (-4), 2 - (-1), 8 - (-2)) = (9, 3, 10)
     \]
   - Vector \( \vec{PR} \) from \( P \) to \( R \):
     \[
     \vec{PR} = R - P = (-1 - (-4), 0 - (-1), 5 - (-2)) = (3, 1, 7)
     \]

2. **Find the Cross Product \( \vec{PQ} \times \vec{PR} \):**
   The cross product of \( \vec{PQ} \) and \( \vec{PR} \) will give us a vector that is perpendicular to the plane, which we can use as the normal vector \( \vec{n} = (a, b, c) \).

   \[
   \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & 3 & 10 \\ 3 & 1 & 7 \end{vmatrix}
   \]
   Expanding this determinant:
   \[
   \vec{PQ} \times \vec{PR} = \hat{i}(3 \cdot 7 - 10 \cdot 1) - \hat{j}(9 \cdot 7 - 10 \cdot 3) + \hat{k}(9 \cdot 1 - 3 \cdot 3)
   \]
   \[
   = \hat{i}(21 - 10) - \hat{j}(63 - 30) + \hat{k}(9 - 9)
   \]
   \[
   = \hat{i}(11) - \hat{j}(33) + \hat{k}(0)
   \]
   \[
   = (11, -33, 0)
   \]

   So, the normal vector \( \vec{n} = (11, -33, 0) \).

3. **Form the Equation of the Plane:**
   Using the point-normal form of the plane, \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \), where \( (x_1, y_1, z_1) \) is a point on the plane (let’s use \( P(-4, -1, -2) \)) and \( \vec{n} = (11, -33, 0) \):
   
   \[
   11(x + 4) - 33(y + 1) + 0(z + 2) = 0
   \]
   Expanding this:
   \[
   11x + 44 - 33y - 33 = 0
   \]
   \[
   11x - 33y + 11 = 0
   \]

4. **Simplify the Equation:**
   Divide through by 11 to simplify:
   \[
   x - 3y + 1 = 0
   \]

Thus, the equation of the plane passing through points \( P \), \( Q \), and \( R \) is:
\[
x - 3y + 1 = 0
\]

Would you like a detailed explanation of each step or have any questions about the process?

Here are five questions to expand your understanding:
1. How can the cross product of two vectors help in determining the equation of a plane?
2. What would change if we used a different point on the plane in the point-normal form?
3. Why is it necessary to simplify the equation after forming it initially?
4. How would you confirm if a fourth point lies on the same plane?
5. Can this approach be extended to higher-dimensional spaces?

**Tip:** Always ensure to check the result by plugging back at least one point into the final equation to verify its correctness.

Find the equation of the plane passing through the points P(-4, -1, -2), Q(5, 2, 8), and R(-1, 0, 5). Write your equation in the form ax + by + cz + d = 0.

Step-by-step solution to find the equation of a plane passing through points P(-4, -1, -2), Q(5, 2, 8), and R(-1, 0, 5) using vector algebra. This guide covers calculating direction vectors, finding the cross product, and deriving the plane equation in standard form. Suitable for high school students.

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