To find the equation of a plane through three points, $(5, -8, -1)$, $(-8, 4, 5)$, and $(-6, -3, 6)$, we will follow these steps:

1. Use the general equation of a plane:

$Ax + By + Cz + D = 0$ where $(A, B, C)$ is the normal vector to the plane.

2. Compute two vectors in the plane:

Let's calculate two vectors from the given points:

• $\vec{v_1} = (-8 - 5, 4 - (-8), 5 - (-1)) = (-13, 12, 6)$
• $\vec{v_2} = (-6 - 5, -3 - (-8), 6 - (-1)) = (-11, 5, 7)$

3. Find the normal vector $\vec{n} = \vec{v_1} \times \vec{v_2}$:

The cross product $\vec{v_1} \times \vec{v_2}$ gives us the normal vector $\vec{n} = (A, B, C)$.

I will calculate the cross product to find the normal vector.The normal vector $\vec{n} = (54, 25, 67)$. So the equation of the plane is:

$54x + 25y + 67z + D = 0$

4. Find $D$:

We substitute the coordinates of one of the given points, say $(5, -8, -1)$, into the plane equation to solve for $D$:

$54(5) + 25(-8) + 67(-1) + D = 0$

Let's calculate $D$.The value of $D$ is $-3$.

Final equation of the plane:

$54x + 25y + 67z - 3 = 0$

This is the equation of the plane passing through the points $(5, -8, -1)$, $(-8, 4, 5)$, and $(-6, -3, 6)$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the angle between two planes given their normal vectors?
2. How can you determine if a point lies on a given plane?
3. What is the geometric significance of the normal vector to a plane?
4. How do you find the distance from a point to a plane?
5. How can we check if two planes are parallel?

Tip: The cross product of two vectors gives a vector that is perpendicular to both, which is useful in finding normals to planes.