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!