We are tasked with finding the point at which the line intersects the plane.

Given:

• Parametric equations of the line: $x = 2 - 2t, \quad y = 3t, \quad z = 1 + t$

• Equation of the plane: $x + 2y - z = 4$

Steps:

1. Substitute the parametric equations of the line into the plane's equation:

   • From the parametric equations, we know: $x = 2 - 2t, \quad y = 3t, \quad z = 1 + t$
   • Substituting these into the plane equation $x + 2y - z = 4$: $(2 - 2t) + 2(3t) - (1 + t) = 4$

2. Simplify the equation: $2 - 2t + 6t - 1 - t = 4$ $(2 - 1) + (-2t + 6t - t) = 4$ $1 + 3t = 4$ $3t = 3$ $t = 1$

3. Find the coordinates of the intersection point: Substitute $t = 1$ back into the parametric equations of the line:

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

   Therefore, the point of intersection is $(0, 3, 2)$.

Conclusion:

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

Would you like more details or have any questions?

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Here are 5 related questions:

1. How do you find the parametric equation of a line passing through two points?
2. What is the geometric interpretation of the intersection of a line and a plane?
3. Can a line be parallel to a plane and not intersect it? Explain.
4. How can you determine if a point lies on a given plane?
5. How can you calculate the distance from a point to a plane?

Tip: Always check if the line is parallel to the plane by comparing their direction vectors. If the dot product of the line's direction vector and the plane's normal vector is zero, the line is parallel to the plane.