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Let's analyze the problem.

### Plane Equation:
The equation of the plane is given by:
\[
\text{Plano: } -x + y + z = 0
\]

### Line Equation:
The parametric equations of the line are:
\[
\text{Reta: } \begin{cases} 
x = 1 + t \\
y = 2 - t \\
z = 5 - t 
\end{cases}
\]

### Analysis of the Statements:

#### I) The line and the plane are parallel.
To check if the line is parallel to the plane, we need to check the dot product between the normal vector of the plane and the direction vector of the line. 

- **Normal vector of the plane**: The plane \( -x + y + z = 0 \) has the normal vector \( \mathbf{n} = (-1, 1, 1) \).
- **Direction vector of the line**: The line's direction vector \( \mathbf{v} \) can be found from the coefficients of \( t \) in the parametric equations, giving us \( \mathbf{v} = (1, -1, -1) \).

Calculate the dot product:
\[
\mathbf{n} \cdot \mathbf{v} = (-1)(1) + (1)(-1) + (1)(-1) = -1 - 1 - 1 = -3 \neq 0
\]
Since the dot product is not zero, the line is **not** parallel to the plane.

#### II) The line and the plane are perpendicular.
The line and the plane are perpendicular if the direction vector of the line is perpendicular to the normal vector of the plane, meaning their dot product should be zero. 

We have already calculated the dot product:
\[
\mathbf{n} \cdot \mathbf{v} = -3
\]
Since the dot product is not zero, the line and the plane are **not** perpendicular.

#### III) The line intersects the plane at the point (3, 0, 3).
To check if the line intersects the plane at the point \( (3, 0, 3) \), we substitute \( x = 3 \), \( y = 0 \), and \( z = 3 \) into the parametric equations and find \( t \).

For \( x = 3 \):
\[
3 = 1 + t \Rightarrow t = 2
\]

For \( y = 0 \) with \( t = 2 \):
\[
y = 2 - t = 2 - 2 = 0
\]

For \( z = 3 \) with \( t = 2 \):
\[
z = 5 - t = 5 - 2 = 3
\]
The point \( (3, 0, 3) \) satisfies the line equations when \( t = 2 \). Let's check if this point satisfies the plane equation:
\[
-3 + 0 + 3 = 0
\]
The point lies on the plane, so the line **does** intersect the plane at \( (3, 0, 3) \).

#### IV) The line has a direction vector \( (1, -1, -1) \).
This statement is correct as we have already identified the direction vector of the line from its parametric equations.

### Conclusion:
The correct options are III and IV.

### Correct Answer:
The correct alternative to select is the one stating that options III and IV are correct.

Considere as equações do plano e da reta definidas abaixo:
Plano: -x + y + z = 0
Reta: x = 1 + t, y = 2 - t, z = 5 - t
Sobre o que foi apresentado afirma-se:
I) A reta e o plano são paralelos.
II) A reta e o plano são perpendiculares.
III) A reta intercepta o plano em (3, 0, 3).
IV) A reta tem vetor diretor (1, -1, -1).
Assinale a alternativa correta.

Learn how to determine the relationship between a line and a plane using parametric equations, vector geometry, and the dot product. This problem involves checking parallelism, perpendicularity, and intersection points. Ideal for Grades 11-12.

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