To find the intersection point between the given line $\mathbf{r}(t) = (-30, -34, 2) + t(-8, -5, 4)$ and the plane $-10x + 2y + 14z = -244$, we can follow these steps:

Step-by-Step Solution

1. Parametric Equation of the Line: The parametric equation of the line $\mathbf{r}(t)$ is given by: $\mathbf{r}(t) = (-30, -34, 2) + t(-8, -5, 4)$ This expands to: $x(t) = -30 - 8t, \quad y(t) = -34 - 5t, \quad z(t) = 2 + 4t$

2. Substitute Parametric Equations into the Plane Equation: The equation of the plane is: $-10x + 2y + 14z = -244$ Now, substitute $x(t), y(t), z(t)$ into this equation: $-10(-30 - 8t) + 2(-34 - 5t) + 14(2 + 4t) = -244$ Simplifying this: $300 + 80t - 68 - 10t + 28 + 56t = -244$ Combine like terms: $300 - 68 + 28 + (80t - 10t + 56t) = -244$ $260 + 126t = -244$ Solve for $t$: $126t = -244 - 260 = -504$ $t = \frac{-504}{126} = -4$

3. Find the Coordinates of the Intersection Point: Substitute $t = -4$ into the parametric equations for $x(t), y(t), z(t)$: $x(-4) = -30 - 8(-4) = -30 + 32 = 2$ $y(-4) = -34 - 5(-4) = -34 + 20 = -14$ $z(-4) = 2 + 4(-4) = 2 - 16 = -14$

Thus, the point of intersection is $(2, -14, -14)$.

Final Answer:

$\boxed{(2, -14, -14)}$

Let me know if you would like further clarification or details on any steps.

Here are five related questions to explore further:

1. How do you find the parametric equation of a line given two points?
2. What happens when the line is parallel to the plane?
3. Can there be cases where a line does not intersect a plane?
4. How can you determine if a line lies entirely in a plane?
5. What is the geometric interpretation of the normal vector to the plane?

Tip: When dealing with parametric equations, always keep track of the parameter values carefully as they help define the specific points on the curve or line.