Final answer:
To find the equation of the plane through two given points and perpendicular to a given plane, we first find the normal vector of the given plane. We can then use one of the given points and the normal vector to find the equation of the desired plane.
Step-by-step explanation:
To find the equation of the plane through the points (2,2,1) and (9,3,6) and perpendicular to the plane 2x + 6y + 6z = 9, we first need to find the normal vector to the given plane. We can do this by finding the coefficients of x, y, and z in the equation. The normal vector is then (2, 6, 6). Now, we can use one of the given points and the normal vector to find the equation of the desired plane.
Let's use the point (2,2,1) and the normal vector (2,6,6). The equation of the plane is given by:
2(x - 2) + 6(y - 2) + 6(z - 1) = 0
Simplifying the equation, we get:
2x + 6y + 6z = 26