Final answer:
The symmetric equations for the line through (2,4,6) that is perpendicular to the plane -3 = 7 are x=2, y=4, and z=6+t.
Step-by-step explanation:
The symmetric equations for a line through a given point and perpendicular to a plane can be found using the normal vector of the plane and the given point. In this case, the equation of the plane is -3 = 7, which has a normal vector of (0, 0, 1).
Since the line is perpendicular to the plane, the direction vector of the line will be parallel to the normal vector of the plane, which is (0, 0, 1).
Using the given point (2, 4, 6) and the direction vector (0, 0, 1), the symmetric equations for the line are x = 2, y = 4, and z = 6 + t, where t is a parameter representing the position along the line.