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To find the equation of the plane that passes through the point \( (1, 1, 1) \) and has a normal vector \( (1, 2, 1) \), we can use the standard form of the equation of a plane:

\[
\vec{n} \cdot (\vec{r} - \vec{r_0}) = 0
\]

Where:
- \( \vec{n} = (1, 2, 1) \) is the normal vector to the plane.
- \( \vec{r} = (x, y, z) \) represents any point on the plane.
- \( \vec{r_0} = (1, 1, 1) \) is a point through which the plane passes (given in the problem).

The equation becomes:
\[
(1, 2, 1) \cdot ((x, y, z) - (1, 1, 1)) = 0
\]
First, subtract the point \( (1, 1, 1) \) from \( (x, y, z) \):
\[
(x - 1, y - 1, z - 1)
\]

Now take the dot product with the normal vector \( (1, 2, 1) \):
\[
1(x - 1) + 2(y - 1) + 1(z - 1) = 0
\]
Simplifying:
\[
(x - 1) + 2(y - 1) + (z - 1) = 0
\]
\[
x - 1 + 2y - 2 + z - 1 = 0
\]
\[
x + 2y + z - 4 = 0
\]

Thus, the equation of the plane is:
\[
x + 2y + z = 4
\]

Now, looking at the answer choices, the correct option corresponds to:
\[
(1, 2, 1) \cdot (\vec{x} - (1, 1, 1)) = 4
\]

So, the correct answer is the third option.

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you find the normal vector of a plane from its equation?
2. Can a plane be uniquely determined by just two points?
3. What happens if the normal vector is perpendicular to the coordinate planes?
4. How can you verify if a point lies on a given plane?
5. What is the geometric meaning of the dot product in plane equations?

**Tip:** A plane's equation can be quickly derived if you know the normal vector and a point on the plane—this is the most efficient way to describe it!

Find an equation of the plane that passes through the point (1, 1, 1) and has normal (1, 2, 1).

Learn how to find the equation of a plane passing through a specific point and having a given normal vector. This problem covers concepts in vector calculus, including plane equations and the dot product. Suitable for high school students in Grades 11-12.

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