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Parametric equations offer a flexible way to describe a line by expressing its coordinates as functions of a parameter, typically denoted as \( t \). This parameter allows us to trace out the position of points along the line as \( t \) varies. For example, consider a line defined by the equations:

• \( x = 3 - 2t \)
• \( y = 1 + 6t \)
• \( z = 2 - \frac{1}{2}t \)

These equations tell us that for any value of \( t \), you can plug it into the equations to find corresponding \( x, y, \) and \( z \) values, thereby obtaining a specific point on the line. In the given problem, you are tasked with determining the value of \( t \) that makes a point on this line also lie on a specified plane, which intersects the space defined by these equations.

A plane in three-dimensional space can be represented by an equation involving a linear combination of \( x, y, \) and \( z \), with constants. The general form of a plane equation is \( ax + by + cz = d \), where \( a, b, \) and \( c \) are coefficients that dictate the orientation of the plane in space. In our exercise, the plane equation is given as:

• \( x + y + 4z = 12 \)

This particular equation establishes a simple flat surface across the space. Each of the coefficients (in this case, 1 for both \( x \) and \( y \), and 4 for \( z \)) helps in determining how the plane tilts relative to the coordinate axes. The constant \( 12 \) locates the position of the plane with respect to the origin; the combination sums the weighted coordinates to equal this constant. Our task involves letting the line intersect this plane, providing a solution where both conditions satisfy each other.

When a plane and a line intersect, determining the specific point of intersection requires a bit of algebraic problem-solving. This process involves substituting the parametric expressions for \( x, y, \) and \( z \) into the plane's equation. Here is how it works:
Starting with the substitutions:

• Plug the line equations \( x = 3 - 2t \), \( y = 1 + 6t \), \( z = 2 - \frac{1}{2}t \) into \( x + y + 4z = 12 \).
• Equation becomes \((3 - 2t) + (1 + 6t) + 4(2 - \frac{1}{2}t) = 12\).

By solving this equation step by step, we simplify it to find \( t \). After substitution and simplification, the equation reduces to \( 12 + 2t = 12 \). Solving for \( t \), we find \( t = 0 \). With this value, we find a specific point on the line that also satisfies the plane's equation.

Understanding the intersection of a line and a plane is crucial in three-dimensional geometry. It demonstrates how objects relate spatially and helps in visualization.
This concept captures how lines can pierce through planes in space. In our problem:

• The line is defined by the movement generated as \( t \) varies in parametric equations.
• The plane creates a slice or a flat sheet in space, acting as a backdrop against which the line moves.
• Intersection is achieved when a point on the line coincides with a point on the plane, determined through solving their equations simultaneously.

Returning to our solution, we established that when \( t = 0 \), the line's point \((3, 1, 2)\) lies on both the line and within the plane, demonstrating their intersection. Exploring such relationships enhances spatial understanding, crucial for fields such as engineering, physics, and computer graphics.