The problem involves finding the equation of a plane passing through a given point $(2, 5, 1)$ with a given normal vector $\mathbf{n} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$.

General Formula for the Plane Equation:

The equation of a plane is given by: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ where $(x_0, y_0, z_0)$ is a point on the plane and $(a, b, c)$ are the components of the normal vector.

Steps to Solve:

1. Use the point $(2, 5, 1)$ as $(x_0, y_0, z_0)$.
2. The components of the normal vector $\mathbf{n} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ are: $a = 1, \, b = -2, \, c = 3$
3. Substituting into the formula: $1(x - 2) - 2(y - 5) + 3(z - 1) = 0$
4. Simplify: $x - 2 - 2y + 10 + 3z - 3 = 0$ $x - 2y + 3z + 5 = 0$

Final Equation of the Plane:

$x - 2y + 3z + 5 = 0$

Let me know if you need further clarification or additional details.

Related Questions:

1. How can you verify that the plane passes through a specific point?
2. What are the properties of the normal vector in the context of planes?
3. How would the equation change if the normal vector was parallel but with different magnitude?
4. How do you determine if a given line lies entirely within this plane?
5. Can this method be used to find the intersection of two planes?

Tip:

When determining the plane equation, ensure the normal vector is not a zero vector, as it would fail to define a unique