If $$A (-1, 4, -3)$$ is one end of a diameter $$AB$$ of the sphere $$x^2 + y^2 + z^2 - 3x - 2y + 2z - 15 = 0$$, then find the co-ordinate of $$B$$.
Given equation of the sphere is $$x^2+y^2+z^2-3 x-2 y+2 z-15=0$$
This is in the form of $$x^2+y^2+z^2+2 a x+2 b y+2 c z+d=0$$
$$\implies a=-\dfrac{3}{2},b=-1,c=1,d =-15$$
Hence the center of the sphere is $$\left(-a,-b,-c\right)=\left(\dfrac{3}{2},1,-1\right)$$
Given $$A(-1,4,-3)$$ is one end of a diameter $$AB$$
Let $$B(x,y,z)$$ be the other end of a diameter
We know that
Center is the mid-point of diameter
$$\implies \dfrac{x-1}{2}=\dfrac{3}{2},\dfrac{y+4}{2}=1,\dfrac{z-3}{2}=-1$$
$$\implies x-1=3,y+4=2,z-3=-2$$
$$\implies x=3+1,y=2-4,z=-2+3$$
$$\implies x=4,y=-2,z=1$$
The co-ordinate of $$B$$ is $$(4,-2,1)$$
Let $$(x,y,z)$$ be points with integer coordinates satisfying the system of homogeneous equations: $$3x-y-z=0$$ $$-3x+z=0$$ $$-3x+2y+z=0$$. Then the number of such points which lie inside a sphere of radius $$10$$ centered at the origin is
The equation of the sphere inscribed in a tetrahedron, whose faces are $$x=0,y=0,z=0$$ and $$x+2y+2z=1$$ is