To find parametric equations for the plane defined by 2x - 3y + z - 6 = 0, the normal vector is identified as (2, -3, 1). Two independent direction vectors in the plane can be derived by finding vectors that are perpendicular to this normal vector. The discussion clarifies that parametric equations for a plane require two parameters, which can be represented using x and y as parameters, leading to z expressed as a function of x and y. A suggested form for the parametric equations is x = u, y = w, and z = 6 - 2u + 3w. This approach simplifies the process of defining the plane parametrically.