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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.

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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.

Find the intersection point(s) (if any) of the line r(t) = (-30, -34, 2) + t(-8, -5, 4) and the plane -10x + 2y + 14z = -244. Point of intersection: ( , , ).

Find the intersection of the line r(t) = (-30, -34, 2) + t(-8, -5, 4) and the plane -10x + 2y + 14z = -244. This problem covers linear algebra concepts like parametric equations and solving systems of equations.

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