A plane containing the point (3,2,0) and the line x=y=z also contains the point (0,7,10), (0,7,1), (0,3,1), (0,-7,1)

Given that point (3,2,0) lies on the plane and line x=y=z also lies on the plane. (1,1,1) are direction ratios of the given line. The plane contains points (3,2,0) and (1,1,1). The direction ratios of the line joining these two points is (2, -1, 1). Therefore, direction ratios of normal to the plane is (i+j+k) x (2i-j+k) = 2i + j -3k. Therefore the equation of the plane is 2x+y-3z=k. Point (3,2,0) lies on the plane, so we get k=8. So the equation of the plane is 2x+y-3z=8. Point (0,7,10) satisfies the plane equation because 2(0)+7-3(10) = 7-30=-23 which is not equal to 8. Point (0,7,1) satisfies the plane equation because 2(0)+7-3(1)=4 which is not equal to 8. Point (0,3,1) satisfies the plane equation because 2(0)+3-3(1)=0 which is not equal to 8. Point (0,-7,1) satisfies the plane equation because 2(0)+(-7)-3(1)=-10 which is not equal to 8. Let's re-examine. The direction ratios of the line x=y=z are (1,1,1). A point on this line is (1,1,1). The vector joining (3,2,0) and (1,1,1) is (1-3, 1-2, 1-0) = (-2,-1,1). The normal vector to the plane is given by the cross product of (-2,-1,1) and (1,1,1): (-2,-1,1) x (1,1,1) = (-2-1, 1-(-2), -2-(-1)) = (-3,3,-1). The equation of the plane is -3x+3y-z=d. Substituting (3,2,0) gives -9+6-0=d, so d=-3. The equation of the plane is -3x+3y-z=-3. Now let's check the points: (0,7,10): -3(0)+3(7)-10=11≠-3 (0,7,1): -3(0)+3(7)-1 = 20≠-3 (0,3,1): -3(0)+3(3)-1 = 8≠-3 (0,-7,1): -3(0)+3(-7)-1=-22≠-3 None of the given points lie on the plane. There might be an error in the problem statement or the given options.