Finding centre of sphere inscribed in tetrahedron Given the tetrahedron with vertices defined by vectors $a=(-4, -3, 1)$, $b=(8,3,1)$, $c= (2, 6, 1)$, $d=(4,3,3)$, find the centre of the sphere inscribed in the tetrahedron.
My train of thought: consider the intersection of the four bisectors of the vertices of the tetrahedron. The centre of the sphere will be in the intersection of the four angle-bisecting planes.
Is this correct?
If so, then we need to find four normal vectors, equations of the four planes, and equate them all together, then find $x$, the centre of the sphere. Is this correct?
Is there a simpler way?
 A: Given any tetrahedron $T$ with vertices $p_1, p_2, p_3, p_4$. Let 


*

*$r$ and $u$ be the in-radius and in-center.

*$(\alpha_1, \alpha_2, \alpha_3, \alpha_4)$ be the barycentric coordinates of $u$ with respect to $T$. i.e. a list of $4$ numbers satisfy:
$$u = \alpha_1 p_1 + \alpha_2 p_2 + \alpha_3 p_3 + \alpha_4 p_4
\quad\text{ and }\quad
\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 1$$

*For each $i$, let $A_i$ be the area of the face of $T$ opposite to $p_i$ and $h_i$ be the height of $p_i$ to that face.

*$V$ be the volume of $T$.


We know that for each $i$, $\displaystyle\;\alpha_i = \frac{r}{h_i}$, together with the fact:
$$
h_1A_1 = h_2A_2 = h_3 A_3 = h_4 A_4 = 3V
$$
We have $\displaystyle\;\alpha_i = \frac{r}{h_i} = \frac{rA_i}{3V}$ and
$\displaystyle\;\sum_{i=1}^4 \alpha_i = 1$ 
reduces to $\displaystyle\;\frac{r}{3V}\sum_{i=1}^4 A_i = 1$.
As a result,
$$u = \sum_{i=1}^4 \alpha_i p_i = \frac{r}{3V}\sum_{i=1}^4 A_i p_i
  = \frac{\sum\limits_{i=1}^4 A_i p_i}{\sum\limits_{i=1}^4 A_i}
$$
i.e the in-center is the area weighted average of the vertices.
The actual computation of the coordinates of in-center for this problem is left as an exercise.
A: The four vertices are $A=(4,3,3),B=(2,6,1),C=(-4,-3,1),D=(8,3,1)$.
The four faces of the tetrahedron are $BCD:z-1=0$; $ACD:x-2y+2z-4=0$; $ABD:x+2y+2z-16=0$; and $ABC:3x-2y-6z+12=0$. So the distance of a point $(a,b,c)$ from the three faces is $|c-1|,\ |\frac{1}{3}|a-2b+2c-4|,\ \frac{1}{3}|a+2b+2c-16|,\ \frac{1}{7}|3a-2b-6c+12|$. These must all be equal.
That gives multiple solutions because we also get all the exspheres. 
So we use JeanMarie's idea that the signs of the distances must be same for the incentre and the centroid which is $G=(\frac{5}{2},\frac{9}{4},\frac{3}{2})$.
The signs of its distances from $BCD,ACD,ABD,ABC$ are +,-,-,+.
So the distances $c-1,-(a-2b+2c-4)/3,-(a+2b+2c-16)/3,(3a-2b-6c+12)/7$ must all be equal.
Solving, we find that $(\frac{26}{7},3,\frac{13}{7})$ is the incentre.
A: You want a point that is equidistant from all 4 planes that make up the tetrahedron.
Find the equations for the 4 planes that are the faces of the sphere.
The find a point that is equidistant from all 4 points.
The formula for the distance of a point from a point from a plane:
If $ax + by + cz - d  = 0$ defines a plane, and $(x_1,y_1,z_1)$ is a point not on the plane then the distance is $\frac{|ax_1+b y_1+c z_1-d|}{\sqrt{a^2+b^2+c^2}}$.
I suggest that when you find the equations for the planes, you normalize them by dividing through by $\sqrt{a^2 + b^2 + c^2}$ 
As JeanMarie points out, there is a sign problem to address.  You could choose equations for the plane such that the normal vectors point toward the center or away from the center.
If you use the distance above (without the absolute values) and find that the distance to the opposite vertex is a positive number, then your normal vectors point in the correct direction. But, if you get a negative number, then multiply through in the equation for the plane by $-1.$
Once you have properly scaled and oriented your equations for the planes, find x,y,z such that:
$(a_1 x + b_1 y + c_1 z - d_1)  = (a_2 x + b_2 y + c_2 z - d_2) = (a_3 x + b_3 y + c_3 z - d_3) =  (a_4 x + b_4 y + c_4 z - d_4)$
