Final answer:
The equation of the plane containing the point (-7, -6, 14) and perpendicular to the y-axis is x = -7. Planes perpendicular to the y-axis have normal vectors parallel to it, such as (0, 1, 0), resulting in an equation with only an x-component.
Step-by-step explanation:
To find an equation of the plane that contains the point (-7, -6, 14) and is perpendicular to the y-axis, we need to understand that a plane perpendicular to the y-axis has a normal vector that is parallel to the y-axis. In this case, a suitable normal vector could be (0, 1, 0) because it only has a component in the y-direction. A general equation of a plane in three-dimensional space can be expressed as Ax + By + Cz + D = 0, where (A, B, C) is the normal vector to the plane.
The equation of the plane with a normal vector parallel to the y-axis would thus have the form x = k, where k is some constant, because the coefficients B and C will be zero since the normal vector does not have x or z components. To determine the value of k, we substitute the point (-7, -6, 14) into the equation, yielding x = -7 as the specific equation of the plane containing the point and perpendicular to the y-axis.