To find the point at which the line intersects the plane, we can proceed as follows:

1. Parametric Equations of the Line:

The parametric equations of the line are given by:

$x = 2 - 2t, \quad y = 3t, \quad z = 1 + t$

2. Equation of the Plane:

The equation of the plane is:

$x + 2y - z = 4$

3. Substituting the Parametric Equations into the Plane Equation:

We will substitute $x = 2 - 2t$, $y = 3t$, and $z = 1 + t$ into the plane equation $x + 2y - z = 4$.

$(2 - 2t) + 2(3t) - (1 + t) = 4$

Simplify the equation:

$2 - 2t + 6t - 1 - t = 4$

$1 + 3t = 4$

4. Solve for $t$:

$3t = 3$

$t = 1$

5. Substitute $t = 1$ back into the parametric equations:

$x = 2 - 2(1) = 0$ $y = 3(1) = 3$ $z = 1 + 1 = 2$

6. Intersection Point:

The point where the line intersects the plane is $(0, 3, 2)$.

Would you like a detailed explanation of any of these steps or further clarification?

Here are 5 questions to expand on this topic:

1. How would the solution change if the plane's equation were different?
2. What if the parametric equation of the line had a different parameter?
3. How can we check if a line lies entirely on the plane?
4. What if the plane's equation was homogeneous (e.g., $x + 2y - z = 0$)?
5. How do you calculate the distance between a point and a plane?

Tip: Always ensure that the parametric equations are correctly substituted into the plane's equation when solving intersection problems.