We are given the parametric equations of the line:

$x = 2 - t, \quad y = 3 + t, \quad z = 4t$

and the equation of the plane:

$x - y + 3z = 9$

To find the point at which the line intersects the plane, we substitute the parametric expressions for $x$, $y$, and $z$ into the plane's equation. This gives us:

$(2 - t) - (3 + t) + 3(4t) = 9$

Now, let's simplify and solve for $t$:

$2 - t - 3 - t + 12t = 9$ $-1 - 2t + 12t = 9$ $-1 + 10t = 9$ $10t = 10$ $t = 1$

Now that we have $t = 1$, we can substitute this value back into the parametric equations to find the corresponding point on the line.

For $x$:

$x = 2 - t = 2 - 1 = 1$

For $y$:

$y = 3 + t = 3 + 1 = 4$

For $z$:

$z = 4t = 4 \times 1 = 4$

So, the point of intersection is:

$(x, y, z) = (1, 4, 4)$

Final Answer: $(x, y, z) = (1, 4, 4)$

Would you like further explanation or details on this solution?

Here are some related questions:

1. How can we find the parametric equation of a line given two points?
2. What is the method for finding the distance from a point to a plane?
3. How can we check if two lines are parallel or intersecting in 3D space?
4. How do we calculate the angle between a line and a plane?
5. What is the concept of a direction vector in the context of parametric equations?

Tip: In problems like this, always make sure to substitute the parametric equations into the plane’s equation and solve for the parameter, as that will give you the point of intersection directly.