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The question asks where the line passing through the points \((-3, 1, 0)\) and \((-1, 3, 4)\) intersects the plane given by the equation \(2x + y - z = -2\).

### Step-by-step solution:

#### Step 1: Find the parametric equation of the line
The line passing through two points can be represented parametrically. If the points are \((-3, 1, 0)\) and \((-1, 3, 4)\), we can write the parametric equations as follows.

Let the parameter \(t\) vary between the two points:

\[
x = -3 + t(-1 - (-3)) = -3 + 2t
\]
\[
y = 1 + t(3 - 1) = 1 + 2t
\]
\[
z = 0 + t(4 - 0) = 4t
\]

Thus, the parametric equation of the line is:
\[
(x, y, z) = (-3 + 2t, 1 + 2t, 4t)
\]

#### Step 2: Find the intersection with the plane
We substitute these parametric equations into the plane equation \(2x + y - z = -2\):

\[
2(-3 + 2t) + (1 + 2t) - 4t = -2
\]

Simplifying this equation:

\[
-6 + 4t + 1 + 2t - 4t = -2
\]
\[
-5 + 2t = -2
\]
\[
2t = 3
\]
\[
t = \frac{3}{2}
\]

#### Step 3: Find the coordinates of the intersection
Now, substitute \(t = \frac{3}{2}\) into the parametric equations for \(x\), \(y\), and \(z\):

\[
x = -3 + 2\left(\frac{3}{2}\right) = -3 + 3 = 0
\]
\[
y = 1 + 2\left(\frac{3}{2}\right) = 1 + 3 = 4
\]
\[
z = 4\left(\frac{3}{2}\right) = 6
\]

Thus, the intersection point is \((0, 4, 6)\).

### Final Answer:
The line intersects the plane at the point \((0, 4, 6)\).

Would you like further details or have any questions about the steps?

Here are 5 related questions:
1. How do we find the direction vector for a line passing through two points?
2. How can we derive the parametric equations of a line from two points?
3. How do you verify that a point lies on a given plane?
4. What is the geometric interpretation of a plane equation like \(2x + y - z = -2\)?
5. How can we use systems of equations to find the intersection of two planes?

**Tip:** The parametric form of a line is often very useful for finding intersections with planes, lines, and other geometric objects!

Where does the line through (-3, 1, 0) and (-1, 3, 4) intersect the plane 2x + y - z = -2?

Find the intersection of a line passing through the points (-3, 1, 0) and (-1, 3, 4) with the plane 2x + y - z = -2. This problem involves parametric equations and solving for the point of intersection, ideal for students in Grades 10-12.

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