The equation of a plane can be written in the general form:

$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$

where $(x_1, y_1, z_1)$ is a point on the plane, and $A, B, C$ are the components of the normal vector to the plane.

Given the point $(6, 2, 9)$ and the normal vector $2\mathbf{i} + 6\mathbf{j} + 5\mathbf{k}$, the components of the normal vector are $A = 2$, $B = 6$, and $C = 5$.

Substitute these values into the general equation:

$2(x - 6) + 6(y - 2) + 5(z - 9) = 0$

Now, simplify the equation:

$2(x - 6) + 6(y - 2) + 5(z - 9) = 0$ $2x - 12 + 6y - 12 + 5z - 45 = 0$ $2x + 6y + 5z - 69 = 0$

So, the equation of the plane is:

$2x + 6y + 5z = 69$

Would you like more details on how this was derived, or do you have any questions?

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Here are five related questions to expand on this topic:

1. How do you determine the normal vector of a plane given three points?
2. What is the geometric interpretation of the normal vector in relation to the plane?
3. How can you determine if a point lies on the plane given its equation?
4. How can you find the angle between two planes given their normal vectors?
5. What is the equation of a plane in parametric form?

Tip: The normal vector defines the orientation of the plane in space and is perpendicular to the plane.