<s>[INST] 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?
 [/INST] 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.
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