Final answer:
The equation of the plane that contains the vectors A, B, and the point C is 6x + 12y + 13z = 39.
Step-by-step explanation:
To find the equation of a plane that contains the vectors A and B, as well as the point C, we can use the cross product of A and B to find the normal vector of the plane. The equation of the plane can be written as Ax + By + Cz = D, where (A, B, C) is the normal vector and (x, y, z) are the coordinates of a point on the plane.
First, find the cross product of vectors A and B:
A x B = (4i + j - 3k) x (i - 3j + 3k)
A x B = (6i + 12j + 13k)
The normal vector of the plane is (6, 12, 13). Now we can substitute the coordinates of point C:
6x + 12y + 13z = D
Substitute the coordinates of point C (2, -1, 3):
6(2) + 12(-1) + 13(3) = D
12 - 12 + 39 = D
D = 39
The equation of the plane that contains the vectors A, B, and the point C is 6x + 12y + 13z = 39.