Write the equation of the plane with a normal vector 
n
=
⟨
1
,
−
5
,
1
⟩
 and the point (1, 2, 3) lying on the plane.

To write the equation of a plane, we use the general formula for the equation of a plane:

\[
Ax + By + Cz = D
\]

Where:
- \(\vec{n} = \langle A, B, C \rangle\) is the normal vector to the plane.
- \((x_1, y_1, z_1)\) is a point on the plane.

Given that the normal vector \(\vec{n} = \langle 1, -5, 1 \rangle\) and the point \((1, 2, 3)\) lies on the plane, we can substitute \(A = 1\), \(B = -5\), and \(C = 1\) into the plane equation. The point \((1, 2, 3)\) will help us find \(D\).

The equation becomes:

\[
1(x - 1) - 5(y - 2) + 1(z - 3) = 0
\]

Now simplify it:

\[
(x - 1) - 5(y - 2) + (z - 3) = 0
\]

Expand the terms:

\[
x - 1 - 5y + 10 + z - 3 = 0
\]

Combine like terms:

\[
x - 5y + z + 6 = 0
\]

Thus, the equation of the plane is:

\[
x - 5y + z + 6 = 0
\]

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you find the normal vector from a given plane equation?
2. What is the equation of a plane parallel to another plane?
3. How can you determine if two planes are perpendicular?
4. What happens when a point is substituted into the plane equation?
5. How do you find the distance between a point and a plane?

**Tip:** Always check that the normal vector is perpendicular to the plane for any given plane equation.

Write the equation of the plane with a normal vector n = ⟨1, −5, 1⟩ and the point (1, 2, 3) lying on the plane.

Learn how to find the equation of a plane using a given normal vector ⟨1, −5, 1⟩ and a point (1, 2, 3) on the plane. This solution explains key concepts in geometry, including the use of vectors and plane equations, suitable for high school students in Grades 10-12.

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Write the equation of the plane with a normal vector n

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