Final answer:
The intersection point of the line and the plane is found by substituting the line's parametric equations into the plane's equation and solving for the parameter. After simplifying the resulting equation, we substitute the parameter back to find the intersection point (3/4, 3, 3/4).
Step-by-step explanation:
To find the point at which the given line intersects the plane defined by the equation x + 2y - z = 6, we need to substitute the parametric equations of the line x = 3 - 3t, y = 4t, z = t into the plane's equation.
First, let's perform the substitution:
- x = 3 - 3t becomes 3 - 3t
- y = 4t becomes 8t (since 2y is in the plane's equation)
- z = t
Now our equation looks like this: 3 - 3t + 8t - t = 6.
Simplify by combining like terms to get 4t = 3, solving for t we find t = 3/4.
Next, substitute t = 3/4 back into the original parametric equations to find the intersection point:
- x = 3 - 3(3/4) = 3/4
- y = 4(3/4) = 3
- z = 3/4
Thus, the intersection point is (3/4, 3, 3/4).