The equation of the sphere inscribed in a tetrahedron, whose faces are $$x=0,y=0,z=0$$ and $$x+2y+2z=1$$ is
Let $$\displaystyle (a,b,c)$$ be the Centre and $$r,$$ the radius of the sphere.
The sphere is inscribed in the tetrahedron, hence the length of the perpendicular from the centre $$(a,b,c)$$ upon each of the faces $$=$$ radius of the sphere
$$\displaystyle \therefore \frac { a }{ 1 } =\frac { b }{ 1 } =\frac { c }{ 1 } =\frac { 1-a-2b-2c }{ \sqrt { 1+4+4 } } =r$$
i.e., $$\displaystyle a=b=c=\frac { 1-a-2b-2c }{ 3 } =r$$ ...(1)
$$\therefore$$ From (1), we get
$$\displaystyle 3a=1-a-2b-2c$$ ...(2)
and, $$\displaystyle a=b=c$$
$$\displaystyle \therefore 3a=1-a-2a-2a\Rightarrow 8a=1$$
$$\displaystyle \therefore a=\frac { 1 }{ 8 } $$
and, then $$\displaystyle r=a=\frac { 1 }{ 8 } $$
$$\therefore$$ Centre is $$\displaystyle \left( \frac { 1 }{ 8 } ,\frac { 1 }{ 8 } ,\frac { 1 }{ 8 } \right) $$ and radius $$\displaystyle =\frac { 1 }{ 8 } $$
Hence, the required sphere is
$$\displaystyle { \left( x-\frac { 1 }{ 8 } \right) }^{ 2 }+{ \left( y-\frac { 1 }{ 8 } \right) }^{ 2 }+{ \left( z-\frac { 1 }{ 8 } \right) }^{ 2 }={ \left( \frac { 1 }{ 8 } \right) }^{ 2 }$$
$$\displaystyle \Rightarrow 32\left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) -8\left( x+y+z \right) +1=0$$
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
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 $$B$$.