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Parametric equations are a way of representing a line using parameters, typically denoted by a variable like \( t \). Each point on the line is represented as a function of \( t \), allowing us to express coordinates \(x, y, z\) in terms of a single parameter. For example, in the line given by \ \( x = 1 - t, \ y = 3t, \ z = 1 + t \), you can imagine \( t \) controlling the position along the line. This parameterization makes it easy to calculate specific points by simply choosing different values of \( t \).
This method is particularly useful because it makes the process of substitution straightforward when solving problems such as finding the intersection of a line with a plane. By substituting the parametric equations into other equations, like the Cartesian equation of a plane, you can solve for \( t \) to find the precise intersection point.

The Cartesian equation of a plane is a linear equation representing a flat surface in three dimensions. This form is typically written as \( ax + by + cz = d \), where \( a, b, \) and \( c \) are coefficients that describe the orientation of the plane, and \( d \) is a constant term.
In the problem, the plane is given by the equation \( 2x - y + 3z = 6 \). In a Cartesian equation, the coefficients \( 2, -1, \) and \( 3 \) correspond to the normal vector of the plane. This is a vector perpendicular to every vector lying on the plane, which defines its orientation in space.
Understanding the Cartesian form of a plane helps in determining how a line will intersect with the plane. To find the intersection, you would substitute the parametric equations of a line into this equation to solve for the specific point.

Solving systems of equations is a crucial step in finding the intersection point between a line and a plane. In this context, you substitute the parametric equations into the plane's Cartesian equation to form a new equation in terms of \( t \). Here, we substitute \( x, y, z \) with \( 1 - t, 3t, 1 + t \) respectively into the equation \( 2x - y + 3z = 6 \), resulting in \( 2(1-t) - 3t + 3(1+t) = 6 \).

After substituting, simplifying the equation involves combining like terms and solving for \( t \), which gives us a specific value that satisfies the equation. In this exercise, solving the system led to \( t = -\frac{1}{2} \).
Finally, by substituting back \( t = -\frac{1}{2} \) into the original parametric equations, the exact intersection point \( \left( \frac{3}{2}, -\frac{3}{2}, \frac{1}{2} \right) \) on the plane is found. This method is a staple technique in various fields requiring linear algebra, offering a systematic way to find where lines and planes meet.