Final answer:
The equation of the plane that passes through (9, 0, −3) and contains the line specified by x = 7 − 2t, y = 3 + 5t, z = 2 + 2t is 14y + 16z + 48 = 0. This is found by using the cross product of the line's direction vector and a vector between the given point and a point on the line to get the plane's normal vector.
Step-by-step explanation:
To find the equation of a plane that passes through a given point and contains a given line, we can use the point on the line and the direction vector of the line, along with the given point that the plane must pass through. The plane that passes through the point (9, 0, −3) and contains the line x = 7 − 2t, y = 3 + 5t, z = 2 + 2t can be found by identifying a point on the line (for example, taking t = 0, we get the point (7, 3, 2)) and the direction vector of the line, which is (−2, 5, 2).
Now, to find the equation of the plane, we can use the cross product of two vectors that lie in the plane. One vector can be formed by finding the displacement between the two points we have: Vector A = (7 − 9, 3 − 0, 2 − (−3)) = (−2, 3, 5). The other vector is the direction vector of the line, which we'll call Vector B = (−2, 5, 2).
Taking the cross product of Vector A and Vector B will give us a normal vector to the plane. Normal vector (N) = Vector A × Vector B = (15 − 15, 4 + 10, 10 + 6) = (0, 14, 16). Finally, using the normal vector and the point (9, 0, −3), we can write the equation of the plane as 0(x − 9) + 14(y − 0) + 16(z + 3) = 0, which simplifies to 14y + 16z + 48 = 0.