In the
plane, we have
everywhere. So in the equation of the sphere, we have

which is a circle centered at (2, -10, 0) of radius 4.
In the
plane, we have
, which gives

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the
plane,
, so

which is a circle centered at (0, -10, 3) of radius
.
I think the answer is 24x-5
The answer is j+5-2
because sum means to add
I think it would be 70382
Answer:
Step-by-step explanation:
4= 0 degree
2z= 1 degree
4r^2st^3= 6 degree
3xyz^2= 4 degree