Final answer:
To find the equation of the plane equidistant from two points, P(1,2,3) and Q(-1,-4,1), find the midpoint of the line segment formed by these two points and the director vector of the line connecting the two points. Use the point-normal form of the equation of a plane to find the equation.
Step-by-step explanation:
To find the equation of the plane equidistant from two points, P(1,2,3) and Q(-1,-4,1), we first need to find the midpoint of the line segment formed by these two points. The midpoint can be found by averaging the x, y, and z coordinates separately. The midpoint is M(0,-1,2).
Next, we find the director vector of the line connecting the two points by subtracting the coordinates of one point from the other. The director vector is D(-2,-6,-2).
Using the point-normal form of the equation of a plane, we can substitute the coordinates of the midpoint, M(0,-1,2), and the direction vector, D(-2,-6,-2), into the equation Ax + By + Cz = -1. This gives us the equation -2x - 6y - 2z = -