If y-intercept of plane \((x-y+z-1)+\lambda(x+y-z-1)=0\) is 3 unit then \(\lambda=\) (A) \(-2\) (B) 2 (C) \((\mathrm{D})-(1 / 2)\)

Understanding 3D geometry is crucial when discussing plane equations and their intercepts. In three-dimensional space, a plane can be thought of as a flat surface extending infinitely in all directions. It's characterized by having length and width but no depth. The intercepts of a plane are the points where the plane crosses the axes of a coordinate system.

For example, a y-intercept occurs where a plane crosses the y-axis. This particular point has an x-coordinate and z-coordinate of zero since it lies directly on the y-axis. When we are given a y-intercept value, such as 3 units in our exercise, it tells us the distance from the origin along the y-axis to the point where the plane intersects.

Important to note is that to find the y-intercept of a plane, you must set the x and z values to zero and solve the resulting equation for the y-coordinate. In three-dimensional geometry, not only can an object have a y-intercept, but it can also have x-intercepts and z-intercepts, which are found in a similar manner by setting the other coordinates to zero.

Moving onto coordinate geometry, this branch of mathematics describes the relationship between geometric figures and algebra using a coordinate system. Here, the position of points on the plane is described using ordered pairs (x, y) in 2D, and triples (x, y, z) in 3D.

Each number in the pair or triple corresponds to a coordinate along a specific axis. For instance, in 3D geometry, these axes are known as the x-axis (horizontal), y-axis (vertical), and z-axis (depth). The coordinates allow us to describe the location or region of geometric figures, such as lines, curves, or planes, in the coordinate system. In the given exercise, the coordinate system enables us to understand the plane's equation in a 3D space and how its y-intercept is related to the variables within the equation.

Moreover, it highlights the significance of each coordinate in determining the specific location of the y-intercept on the plane. Algebraically, finding the y-intercept necessitates setting the x and z coordinates to zero, because the y-intercept is where the plane crosses the y-axis.

Lastly, vector algebra is relevant when dealing with directions and magnitudes in 3D geometry. It involves operations with vectors, which are quantities that have both a magnitude and a direction. The plane's normal vector, for instance, is a vector that is perpendicular to the surface of the plane.

In our exercise, the equation \[ (x-y+z-1)+\lambda(x+y-z-1)=0 \] implicitly defines such a vector, since the coefficients of x, y, and z give its direction. As parameter \( \lambda \) changes, it alters the coefficients, essentially rotating or sliding the plane around the 3D space without changing its 'slope', revealing different y-intercepts. This highlights how vector algebra can dynamically represent geometric objects and showcases the relationship between algebraic parameters and geometric properties.

Algebraic manipulation within vector algebra can lead to solving for unknown parameters, such as the \( \lambda \) in our exercise's equation. This scalar determines the influence of the additional vector on the position of the plane. By understanding these vector operations, we can solve complex 3D geometry problems by treating them as algebraic expressions.