Final answer:
To find the equation of the plane that passes through the given points and is perpendicular to the given plane equation, we can find the direction vector of the plane and use it along with one of the points to obtain the equation of the plane.
Step-by-step explanation:
To find the equation of the plane that passes through the given points and is perpendicular to the plane equation, we first need to find the direction vector of the plane. The direction vector is the normal vector to the plane, which can be found by taking the coefficients of x, y, and z in the given plane equation. For the equation 7x + 8y + 3z = 16, the direction vector is (7, 8, 3).
Next, we can use the two given points to find the position vector of the plane. The position vector is the vector that connects one of the points to any point on the plane. In this case, we can choose the first point (3, 2, 1). So the position vector is (3-3, 2-2, 1-1) = (0, 0, 0).
Finally, we can use the direction vector and the position vector to obtain the equation of the plane. The equation of a plane is given by Ax + By + Cz = D, where (A, B, C) is the direction vector and (x, y, z) is the position vector. Plugging in the values, we get 7(x-3) + 8(y-2) + 3(z-1) = 0, which simplifies to 7x + 8y + 3z = 37.